C3 Edexcel Solution Bank - Chapter 7 - BioChem Tuition · Solutionbank Edexcel AS and A Level Modular Mathematics Exercise A, Question 4 Question: Express the following as a single

Solutionbank Edexcel AS and A Level Modular Mathematics Exercise A, Question 1

© Pearson Education Ltd 2008

Question:

A student makes the mistake of thinking that sin (A + B ) ≡ sinA + sinB . Choose non-zero values of A and B to show that this statement is not true for all values of A and B.

Solution:

Example: Take A = 30 ° , B = 60 °

sinA =

sinB =

sinA + sinB ≠ 1 but sin (A + B ) = sin90 ° = 1 . This proves that sin (A + B ) = sinA + sinB is not true for all values. There will be many values of A and B for which it is true, e.g. A = − 30 ° and B = + 30 ° , and that is the danger of trying to prove a statement by taking particular examples. To prove a statement requires a sound argument; to disprove only requires one example.

1

2

√ 3

2

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Question:

Using the expansion of cos (A − B ) with A = B = θ, show that sin2

θ + cos2 θ ≡ 1.

Solution:

cos (A − B ) ≡ cosAcosB + sinAsinB Set A = θ, B = θ ⇒ cos (θ − θ ) ≡ cosθcosθ + sinθsinθ

⇒ cos0≡ cos2 θ + sin2θ

So cos2 θ + sin2θ ≡ 1 (since cos0 = 1)


3/9/2013file://C:\Users\Buba\kaz\ouba\c3_7_a_2.html




Question:

(a) Use the expansion of sin (A − B ) to show that sin −θ = cosθ .

(b) Use the expansion of cos (A − B ) to show that cos −θ = sinθ .

π

2

π

2

Solution:

(a) sin (A − B ) ≡ sinAcosB − cosAsinB

Set A = , B = θ

⇒ sin − θ ≡ sin cosθ − cos sinθ

⇒ sin − θ ≡ cosθ (since sin = 1 , cos = 0)

(b) cos (A − B ) ≡ cosAcosB + sinAsinB

Set A = , B = θ

⇒ cos −θ ≡ cos cosθ + sin sinθ

⇒ cos −θ ≡ sinθ (since cos = 0 , sin = 1)

π

2

π

2

π

2

π

2

π

2

π

2

π

2

π

2

π

2

π

2

π

2

π

2

π

2

π

2


3/9/2013



Question:

Express the following as a single sine, cosine or tangent:

(a) sin15 ° cos20 ° + cos15 ° sin20 °

(b) sin58 ° cos23 ° − cos58 ° sin23 °

(c) cos130 ° cos80 ° − sin130 ° sin80 °

(d)

(e) cos2θcosθ + sin2θsinθ

(f) cos4θcos3θ − sin4θsin3θ

(g) sin θcos2 θ + cos θsin2 θ

(h)

(i) sin ( A + B ) cosB − cos (A + B ) sinB

(j) cos cos − sin sin

tan76 ° − tan45 °

1 + tan76 ° tan45 °

1

2

1

2

1

2

1

2

tan2θ+ tan3θ

1 − tan2θtan3θ

3x + 2y

2

3x − 2y

2

3x + 2y

2

3x − 2y

2

Solution:

(a) Using sin (A + B ) ≡ sinAcosB + cosAsinB sin15 ° cos20 ° + cos15 ° sin20 °≡ sin ( 15 ° + 20 ° ) ≡ sin35 °

(b) Using sin (A − B ) ≡ sinAcosB − cosAsinB sin58 ° cos23 ° − cos58 ° sin23 °≡ sin ( 58 ° − 23 ° ) ≡ sin35 °

(c) Using cos (A + B ) ≡ cosAcosB − sinAsinB cos130 ° cos80 ° − sin130 ° sin80 °≡ cos ( 130 ° + 80 ° )≡ cos210 °

(d) Using tan (A − B ) ≡tanA − tanB

1 + tanA tanB





≡ tan ( 76 ° − 45 ° ) ≡ tan31 °

(e) Using cos (A − B ) ≡ cosAcosB + sinAsinB cos2θcosθ + sin2θsinθ ≡ cos ( 2θ − θ ) ≡ cosθ

(f) Using cos (A + B ) ≡ cosAcosB − sinAsinB cos4θcos3θ − sin4θsin3θ ≡ cos ( 4θ + 3θ ) ≡ cos7θ

(g) Using sin (A + B ) ≡ sinAcosB + cosAsinB

sin θcos2 θ + cos θsin2 θ ≡ sin θ + 2 θ ≡ sin3θ

(h) Using tan (A + B ) ≡

≡ tan ( 2θ + 3θ ) ≡ tan5θ

(i) Using sin (P − Q ) ≡ sinPcosQ − cosPsinQ sin ( A + B ) cosB − cos (A + B ) sinB ≡ sin [ ( A + B ) − B ] ≡ sinA

(j) Using cos (A + B ) ≡ cosAcosB − sinAsinB

cos cos − sin sin

≡ cos + ≡ cos ≡ cos3x

tan76 ° − tan45 °

1 + tan76 ° tan45 °

1

2

1

2

1

2

1

2

1

2

1

2

tanA + tanB

1 − tanA tanB

tan2θ+ tan3θ

1 − tan2θtan3θ

3x + 2y

2

3x − 2y

2

3x + 2y

2

3x − 2y

2

3x + 2y

2

3x − 2y

2

6x

2





Question:

Calculate, without using your calculator, the exact value of:

(a) sin30 ° cos60 ° + cos30 ° sin60 °

(b) cos110 ° cos20 ° + sin110 ° sin20 °

(c) sin33 ° cos27 ° + cos33 ° sin27 °

(d) cos cos − sin sin

(e) sin60 ° cos15 ° − cos60 ° sin15 °

(f) cos70 ° ( cos50 ° − tan70 ° sin50 ° )

(g)

(h)

(i)

(j) √ 3cos15 ° − sin15 °

π

8

π

8

π

8

π

8

tan45 ° + tan15 °

1 − tan45 ° tan15 °

1 − tan15 °

1 + tan15 °

tan ( ) − tan ( )7π

12

π

3

1 + tan ( ) tan ( )7π

12

π

3

Solution:

(a) Using sin (A + B ) expansion sin30 ° cos60 ° + cos30 ° sin60 ° = sin ( 30 + 60 ) ° = sin90 ° = 1

(b) cos110 ° cos20 ° + sin110 ° sin20 ° = cos ( 110 − 20 ) ° = cos90 ° = 0

(c) sin33 ° cos27 ° + cos33 ° sin27 ° = sin ( 33 + 27 ) ° = sin60 ° =

(d) cos cos − sin sin = cos + = cos =

(e) sin60 ° cos15 ° − cos60 ° sin15 ° = sin ( 60 − 15 ) ° = sin45 ° =

√ 3

2

π

8

π

8

π

8

π

8

π

8

π

8

π

4

√ 2

2

√ 2

2





(f) cos70 ° cos50 ° − cos70 ° tan70 ° sin50 °= cos70 ° cos50 ° − sin70 ° sin50 °

= cos ( 70 + 50 ) °

= cos120 °

= − cos60 ° = −

(g) = tan ( 45 + 15 ) ° = tan60 ° =√ 3

(h) = (using tan45 ° = 1)

= tan ( 45 − 15 ) ° = tan30 ° =

(i) = tan − = tan = tan = 1

(j) This is very similar to part (e) but you need to rewrite it as 2 cos15 ° −

sin15 ° to appreciate it!

√ 3cos15 ° − sin15 °≡ 2 cos15 ° − sin15 °

≡ 2 ( sin60 ° cos15 ° − cos60 ° sin15 ° )≡ 2sin ( 60 − 15 ) °

≡ 2sin45 ° = √ 2

1

2

tan45 ° + tan15 °

1 − tan45 ° tan15 °

1 − tan15 °

1 + tan15 °

tan45 ° − tan15 °

1 + tan45 ° tan15 °

√ 3

3

tan ( ) − tan ( π )7π

12

1

3

1 + tan ( ) tan ( π )7π

12

1

3

7π

12

π

3

3π

12

π

4

√ 3

2

1

2

√ 3

2

1

2





Question:

Triangle ABC is such that AB = 3cm, BC = 4cm, ∠ ABC = 120 ° and

∠ BAC = θ ° .

(a) Write down, in terms of θ, an expression for ∠ACB.

(b) Using the sine rule, or otherwise, show that tanθ ° = .2 √ 3

5

Solution:

(a) ∠ ACB = 180 ° − 120 ° −θ ° = ( 60 −θ ) °

(b) Using sine rule: =

⇒ =

⇒ 4sin ( 60 −θ ) ° = 3sinθ °

⇒ 4sin60 ° cosθ ° − 4cos60 ° sinθ ° = 3sinθ °

⇒ 2 √ 3cosθ ° − 2sinθ ° = 3sinθ ° sin60 ° =

⇒ 5sinθ ° = 2 √ 3cosθ °

sinC

c

sinA

a

sin ( 60 −θ ) °

3

sinθ °

4

√ 3

2





⇒ =

⇒ tanθ ° =

sinθ °

cosθ °

2 √ 3

5

2 √ 3

5





Question:

Prove the identities

(a) sin ( A + 60 ° ) + sin (A − 60 ° ) ≡ sinA

(b) − ≡

(c) ≡ tanx + tany

(d) + 1 ≡ cot xcot y

(e) cos θ + + √ 3sinθ ≡ sin θ +

(f) cot ( A + B ) ≡

(g) sin2 ( 45 + θ ) ° + sin2 ( 45 − θ ) ° ≡ 1

(h) cos ( A + B ) cos ( A − B ) ≡ cos2 A − sin2 B

cosA

sinB

sinA

cosB

cos (A + B )

sinBcosB

sin ( x + y )

cosxcosy

cos (x + y )

sinxsiny

π

3

π

6

cot Acot B − 1

cot A + cot B

Solution:

(a) L.H.S. ≡ sin ( A + 60 ° ) + sin (A − 60 ° )≡ sinAcos60 ° + cosAsin60 ° + sinAcos60 ° − cosAsin60 °≡ 2sinAcos60 °

≡ sinA (since cos60 ° = )

≡ R.H.S.

(b) L.H.S. ≡ −

≡

≡

≡ R.H.S.

(c) L.H.S. ≡

1

2

cosA

sinB

sinA

cosB

cosAcosB − sinAsinB

sinBcosB

cos (A + B )

sinBcosB

sin ( x + y )

cosxcosy




≡

≡ +

≡ +

≡ tanx + tany ≡ R.H.S.

(d) L.H.S. ≡ + 1

≡

≡

≡

≡ cot xcot y ≡ R.H.S.

(e) L.H.S. ≡ cos θ + + √ 3sinθ

≡ cosθcos − sinθsin + √ 3sinθ

≡ cosθ − sinθ + √ 3sinθ

≡ sinθ + cosθ

≡ sinθcos + cosθsin cos = , sin =

≡ sin θ + [ sin ( A + B ) ]

≡ R.H.S.

(f)

sinxcosy + cosx siny

cosx cosy

sinxcosy

cosx cosy

cosxsiny

cosxcosy

sinx

cosx

siny

cosy

cos (x + y )

sinxsiny

cos (x + y ) + sinx siny

sinx siny

cosx cosy − sinx siny + sinx siny

sinxsiny

cosx cosy

sinx siny

π

3

π

3

π

3

1

2

√ 3

2

√ 3

2

1

2

π

6

π

6

π

6

√ 3

2

π

6

1

2

π

6




(g) L.H.S. ≡ sin2 ( 45 + θ ) ° + sin2 ( 45 − θ ) °≡ ( sin45 ° cosθ ° + cos45 ° sinθ ° ) 2 + ( sin45 ° cosθ ° − cos45 °

sinθ ° ) 2 As sin45 ° = cos45 ° it is easier to take out as a common factor. ≡ ( sin45 ° ) 2 [ ( cosθ ° + sinθ ° ) 2 + ( cosθ ° − sinθ ° ) 2 ]

≡ cos2 θ ° + 2sinθ ° cosθ ° + sin2θ ° + cos2 θ ° − 2sinθ °

cosθ ° + sin2θ °

≡ 2 sin2 θ ° + cos2 θ °

≡ × 2 ( sin2 θ ° + cos2 θ ° ≡ 1 )

≡ 1 ≡ R.H.S.

Alternatively: as sin ( 90 ° −x ° ) ≡ cosx ° , if x = 45 ° + θ ° then sin ( 45 ° −θ ° ) ≡ cos ( 45 ° +θ ° )and original L.H.S. becomes sin2 ( 45 + θ ) ° + cos2 ( 45 + θ ) °which is identically = 1

(h) L.H.S. ≡ cos (A + B ) cos (A − B )≡ ( cosAcosB − sinAsinB ) ( cosAcosB + sinAsinB )≡ cos2 Acos2 B − sin2 Asin2 B

1

2

1

2

1

2





≡ cos2 A ( 1 − sin2 B ) − ( 1 − cos2 A ) sin2 B

≡ cos2 A − cos2 Asin2 B − sin2 B + cos2 Asin2 B

≡ cos2 A − sin2 B ≡ R.H.S.






Question:

Given that sinA = and sinB = , where A and B are both acute angles, calculate the

exact values of

(a) sin ( A + B )

(b) cos (A − B )

(c) sec ( A − B )

4

5

1

2

Solution:

Using Pythagoras' theorem x = 3 and y = √ 3

(a) sin ( A + B ) = sinAcosB + cosAsinB = × + × =

(b) cos (A − B ) = cosAcosB + sinAsinB = × + × =

(c) sec (A − B ) = =

=

=

=

4

5

√ 3

2

3

5

1

2

4 √ 3 + 3

10

3

5

√ 3

2

4

5

1

2

3 √ 3 + 4

10

1

cos (A − B )

10

3 √ 3 + 4

10 ( 3√ 3 − 4 )

( 3 √ 3 + 4 ) ( 3√ 3 − 4 )

10 ( 3√ 3 − 4 )

27 − 16

10 ( 3√ 3 − 4 )

11





Question:

Given that cosA = − , and A is an obtuse angle measured in radians, find the

exact value of

(a) sin A

(b) cos (π + A )

(c) sin +A

(d) tan +A

4

5

π

3

π

4

Solution:

Draw a right-angled triangle where cosA′ =

Using Pythagoras' theorem x = 3

So sinA′ = , tanA′ =

(a) As A is in the 2nd quadrant, sinA = sinA′

sinA =

(b) cos (π + A ) = cosπcosA − sinπsinA = − cosA ( cosπ = − 1 , sinπ = 0 )

4

5

3

5

3

4

3

5





cos (π + A ) = +

(c) sin +A = sin cosA + cos sinA

= − +

=

(d) As A is in 2nd quadrant, tanA = − tanA′ = −

tan +A = = = =

4

5

π

3

π

3

π

3

√ 3

2

4

5

1

2

3

5

3 − 4 √ 3

10

3

4

π

4

tan + tanAπ

4

1 − tan tanAπ

4

1 + tanA

1 − tanA

1

4

7

4

1

7





Question:

Given that sinA = , where A is acute, and cosB = − , where B is obtuse,

calculate the exact value of

(a) sin (A − B )

(b) cos (A − B )

(c) cot ( A − B )

8

17

4

5

Solution:

sinB = sinB′, tanB = − tanB′ By Pythagoras' theorem, the remaining sides are 15 and 3.

So sinA = , cosA = , tanA =

and sinB = , cosB = − , tanB = −

(a) sin (A − B ) = sinAcosB − cosAsinB

= − −

= = −

(b) cos (A − B ) = cosAcosB + sinAsinB

= − +

8

17

15

17

8

15

3

5

4

5

3

4

8

17

4

5

15

17

3

5

− 32 − 45

85

77

85

15

17

4

5

8

17

3

5





= = −

(c) tan (A − B ) = = = =

So cot (A − B ) = =

− 60 + 24

85

36

85

tanA − tanB

1 + tanA tanB

+8

15

3

4

1 −24

60

77

60

36

60

77

36

1

tan (A − B )

36

77





Question:

Given that tanA = , where A is reflex, and sinB = , where B is obtuse,

calculate the exact value of

(a) sin (A + B )

(b) tan (A − B )

(c) cosec( A + B )

7

24

5

13

Solution:

Using Pythagoras' theorem, the remaining sides are 25 and 12. As A is in the 3rd quadrant (tan A is +ve, and A is reflex), sinA = − sinA′, cosA = − cosA′

So sinA = − , cosA = − , tanA =

As B is in the 2nd quadrant, cosB = − cosB′, tanB = − tanB′

So sinB = , cosB = − , tanB = −

(a)

7

25

24

25

7

24

5

13

12

13

5

12

sin ( A + B ) = sinAcosB + cosAsinB

= ( − ) ( − ) + ( − ) ( )7

25

12

13

24

25

5

13

= = −84 − 120

325

36

325





(b) tan (A − B ) = = = =

(c) cosec (A + B ) = = −

tanA − tanB

1 + tanA tanB

+7

24

5

12

1 − ( ) ( )7

24

5

12

17

24

253

288

204

253

1

sin ( A + B )

325

36





Question:

Write the following as a single trigonometric function, assuming that θ is measured in radians:

(a) cos2 θ − sin2θ

(b) 2sin4θcos4θ

(c)

(d) ( sinθ + cosθ )

1 + tanθ

1 − tanθ

1

√ 2

Solution:

(a) cos2 θ − sin2θ = cosθcosθ − sinθsinθ = cos (θ + θ ) = cos2θ

(b) 2sin4θcos4θ = sin4θcos4θ + sin4θcos4θ = sin4θcos4θ + cos4θsin4θ = sin ( 4θ + 4θ ) = sin8θ

(c) = (as tan = 1)

= tan +θ

(d) sinθ + cosθ = sinθcos + cosθsin (as = cos = sin

)

= sin θ +

1 + tanθ

1 − tanθ

tan + tanθπ

4

1 − tan tanθπ

4

π

4

π

4

1

√ 2

1

√ 2

π

4

π

4

1

√ 2

π

4

π

4

π

4





[ Note: (d) could be cos θ −

π

4





Question:

Solve, in the interval 0 ° ≤ θ < 360 ° , the following equations. Give answers to the nearest 0.1°.

(a) 3cosθ = 2sin (θ + 60 ° )

(b) sin (θ + 30 ° ) + 2sinθ = 0

(c) cos (θ + 25 ° ) + sin (θ + 65 ° ) = 1

(d) cosθ = cos (θ + 60 ° )

(e) tan (θ − 45 ° ) = 6 tanθ

(f) sinθ + cosθ = 1

Solution:

(a) 3cosθ = 2sin (θ + 60 ° )⇒ 3cosθ = 2 ( sinθcos60 ° + cosθsin60 ° )

⇒ 3cosθ = 2 sinθ + cosθ = sinθ + √ 3cosθ

⇒ ( 3 − √ 3 ) cosθ = sinθ

⇒ tanθ = 3 − √ 3 tanθ =

As tan θ is +ve, θ is in 1st and 3rd quadrants. θ = tan− 1 ( 3 − √ 3 ) , 180 ° + tan− 1 ( 3 − √ 3 ) = 51.7°, 231.7°

(b) sin (θ + 30 ° ) + 2sinθ = 0 ⇒ sinθcos30 ° + cosθsin30 ° + 2sinθ = 0

⇒ sinθ + cosθ + 2sinθ = 0

⇒ ( 4 + √ 3 ) sinθ = − cosθ

⇒ tanθ = −

As tan θ is −ve, θ is in the 2nd and 4th quadrants.

1

2

√ 3

2

sinθ

cosθ

√ 3

2

1

2

1

4 + √ 3




θ = tan− 1 − + 180 ° , tan− 1 − + 360 °

= 170.1 ° , 350.1°.

(c) cos (θ + 25 ° ) + sin (θ + 65 ° ) = 1 ⇒ cosθcos25 ° − sinθsin25 ° + sinθcos65 ° + cosθsin65 ° = 1

As sin ( 90 −x ) ° = cosx ° and cos ( 90 −x ) ° = sinx ° cos25 ° = sin65 ° and sin25 ° = cos65 ° So cosθsin65 ° − sinθcos65 ° + sinθcos65 ° + cosθsin65 ° = 1 ⇒ 2cosθsin65 ° = 1

⇒ cosθ = = 0.55169

θ = cos− 1 ( 0.55169 ) , 360 ° − cos− 1 ( 0.55169 ) = 56.5°, 303.5°

(d) cosθ = cos (θ + 60 ° )

⇒ cosθ = cosθcos60 ° − sinθsin60 ° = cosθ − sinθ

⇒ cosθ = − √ 3sinθ

⇒ tanθ = − tanθ =

θ is in the 2nd and 4th quadrants.

1

4 + √ 3

1

4 + √ 3

1

2sin 65 °

1

2

√ 3

2

1

√ 3

sinθ

cosθ




tan− 1 − is not in given interval.

θ = tan− 1 − + 180 ° , tan− 1 − + 360 ° = 150 ° ,

330°

(e) tan (θ − 45 ° ) = 6 tanθ

⇒ = 6tanθ

⇒ = 6tanθ

⇒ tanθ − 1 = 6tanθ + 6 tan2 θ

⇒ 6tan2 θ + 5tanθ + 1 = 0

⇒ ( 3 tanθ + 1 ) ( 2 tanθ + 1 ) = 0

⇒ tanθ = − or tanθ = −

tanθ = − ⇒ θ = tan− 1 − + 180 ° , tan− 1 −

+ 360 ° = 161.6 ° , 341.6°

tanθ = − ⇒ θ = tan− 1 − + 180 ° , tan− 1 −

+ 360 ° = 153.4 ° , 333.4° Set of solutions: 153.4°, 161.6°, 333.4°, 341.6°

(f) sinθ + cosθ = 1

⇒ sinθ + cosθ =

1

√ 3

1

√ 3

1

√ 3

tanθ − tan 45 °

1 + tanθ tan45 °

tanθ − 1

1 + tanθ

1

3

1

2

1

3

1

3

1

3

1

2

1

2

1

2

1

√ 2

1

√ 2

1

√ 2





⇒ cos45 ° sinθ + sin45 ° cosθ =

⇒ sin ( θ + 45 ° ) =

⇒ θ + 45 ° = 45 ° , 135°

⇒ θ = 0 ° , 90°

1

√ 2

1

√ 2






Question:

(a) Solve the equation cosθcos30 ° − sinθsin30 ° = 0.5, for 0 ≤ θ ≤ 360 ° .

(b) Hence write down, in the same interval, the solutions of √ 3cosθ − sinθ = 1.

Solution:

(a) cosθcos30 ° − sinθsin30 ° = 0.5 ⇒ cos (θ + 30 ° ) = 0.5

⇒ θ + 30 ° = 60 ° , 300°

⇒ θ = 30 ° , 270°

(b) cosθcos30 ° − sinθsin30 ° ≡ cosθ × − sinθ ×

So cosθcos30 ° − sinθsin30 ° =

is identical to √ 3cosθ − sinθ = 1 Solutions are same as (a), i.e. 30°, 270°

√ 3

2

1

2

1

2







Question:

(a) Express tan ( 45 + 30 ) ° in terms of tan 45° and tan 30°.

(b) Hence show that tan75 ° = 2 + √ 3.

Solution:

(a) tan ( 45 + 30 ) ° =

(b)

=

=

=

= 2 + √ 3

tan45 ° + tan30 °

1 − tan45 ° tan30 °

√ 3 + 1

√ 3 − 1

( √ 3 + 1 ) ( √ 3 + 1 )

( √ 3 − 1 ) ( √ 3 + 1 )

4 + 2 √ 3

2







Question:

Show that sec105 ° = − √ 2 ( 1 + √ 3 )

Solution:

cos ( 60 + 45 ) ° = cos60 ° cos45 ° − sin60 ° sin45 °

= × − ×1

2

1

√ 2

√ 3

2

1

√ 2

=1 − √ 3

2 √ 2

So sec105 ° = =1

cos105 °

2 √ 2

1 − √ 3

=2 √ 2 ( 1 + √ 3 )

( 1 − √ 3 ) ( 1 + √ 3 )

=2 √ 2 ( 1 + √ 3 )

− 2

= − √ 2 ( 1 + √ 3 )


3/9/2013file://C:\Users\Buba\kaz\ouba\c3_7_a_16.htm



Question:

Calculate the exact values of

(a) cos 15°

(b) sin 75°

(c) sin ( 120 + 45 ) °

(d) tan 165°

Solution:

(a) cos15 ° = cos ( 45 − 30 ) °

= cos45 ° cos30 ° + sin45 ° sin30 °

= × + ×

= ( √ 3 + 1 )

(b) sin75 ° = sin ( 45 + 30 ) °

= sin45 ° cos30 ° + cos45 ° sin30 °

= × + ×

= ( √ 3 + 1 )

[ sin75 ° = cos ( 90 − 75 ° ) = cos15 ° ]

(c) sin ( 120 + 45 ) ° = sin120 ° cos45 ° + cos120 ° sin45 °

= × + −

= ( √ 3 − 1 )

(d) tan165 ° = tan ( 120 + 45 ) ° = ( tan120 °

= − √ 3 )

√ 2

2

√ 3

2

√ 2

2

1

2

√ 2

4

√ 2

2

√ 3

2

√ 2

2

1

2

√ 2

4

√ 3

2

√ 2

2

1

2

√ 2

2

√ 2

4

tan120 ° + tan45 °

1 − tan120 ° tan45 °





=− √ 3 + 1

1 + √ 3

=( − √ 3 + 1 ) ( √ 3 − 1 )

( √ 3 + 1 ) ( √ 3 − 1 )

=− 4 + 2√ 3

2

= − 2 + √ 3






Question:

(a) Given that 3sin (x − y ) − sin ( x + y ) = 0 , show that tanx = 2tany.

(b) Solve 3sin ( x − 45 ° ) − sin ( x + 45 ° ) = 0 , for 0 ≤ x ≤ 360 ° .

Solution:

(a) 3sin (x − y ) − sin ( x + y ) = 0 ⇒ 3sinx cosy − 3cosxsiny − sinxcosy − cosxsiny = 0

⇒ 2sinx cosy = 4cosxsiny

⇒ =

⇒ =

⇒ 2tanx = 4tanySo tanx = 2tany

(b) Put y = 45 ° ⇒ tanx = 2

So x = tan− 1 2, 180 ° + tan− 1 2 = 63.4 ° , 243.4°

2sinxcosy

cosxcosy

4cosxsiny

cosxcosy

2sinx

cosx

4siny

cosy






Question:

Given that sinx ( cosy + 2siny ) = cosx ( 2cosy − siny ) , find the value of tan ( x + y ) .

Solution:

sinx ( cosy + 2siny ) = cosx ( 2cosy − siny )⇒ sinxcosy + 2sinxsiny = 2cosxcosy − cosxsiny

⇒ sinxcosy + cosxsiny = 2 ( cosxcosy − sinxsiny )

⇒ sin ( x + y ) = 2cos (x + y )

⇒ = 2

⇒ tan (x + y ) = 2

sin ( x + y )

cos (x + y )






Question:

Given that tan ( x − y ) = 3 , express tan y in terms of tan x.

Solution:

As tan (x − y ) = 3

so = 3

⇒ tanx − tany = 3 + 3 tanx tany

⇒ 3tanx tany + tany = tanx − 3

⇒ tany ( 3 tanx + 1 ) = tanx − 3

⇒ tany =

tanx − tany

1 + tanx tany

tanx − 3

3 tanx + 1





Question:

In each of the following, calculate the exact value of tan x°.

(a) tan (x − 45 ) ° =

(b) sin (x − 60 ) ° = 3cos (x + 30 ) °

(c) tan ( x − 60 ) ° = 2

1

4

Solution:

(a) tan (x − 45 ) ° =

⇒ =

⇒ 4tanx ° − 4 = 1 + tanx ° ( tan45 ° = 1 )

⇒ 3tanx ° = 5

⇒ tanx ° =

(b) sin (x − 60 ) ° = 3cos (x + 30 ) °

⇒ sinx ° cos60 ° − cosx ° sin60 ° = 3cosx ° cos30 ° − 3sinx °sin30 °

⇒ sinx ° × − cosx ° × = 3cosx ° × − 3sinx ° ×

⇒ 4sinx ° = 4 √ 3cosx °

⇒ =

⇒ tanx ° = √ 3

(c) tan (x − 60 ) ° = 2

⇒ = 2

1

4

tanx ° − tan45 °

1 + tanx ° tan45 °

1

4

5

3

1

2

√ 3

2

√ 3

2

1

2

sinx °

cosx °

4 √ 3

4

tanx ° − tan60 °

1 + tanx ° tan60 °





⇒ = 2

⇒ tanx ° − √ 3 = 2 + 2√ 3tanx °

⇒ ( 2 √ 3 − 1 ) tanx ° = − ( 2 + √ 3 )

⇒ tanx ° = − = − = −

tanx ° − √ 3

1 + √ 3tanx °

( 2 + √ 3 )

2 √ 3 − 1

( 2 + √ 3 ) ( 2 √ 3 + 1 )

( 2 √ 3 − 1 ) ( 2√ 3 + 1 )

8 + 5 √ 3

11






Question:

Given that tanA ° = and tanB ° = , calculate, without using your calculator, the

value of A + B,

(a) where A and B are both acute,

(b) where A is reflex and B is acute.

1

5

2

3

Solution:

(a) tan (A ° + B ° ) =

As tan (A + B ) ° is +ve, A + B is in the 1st or 3rd quadrants, but as they are both acute A + B cannot be in the 3rd quadrant. So (A + B ) ° = tan− 1 1 = 45 °i.e. A + B = 45

(b)A is reflex but tanA ° is +ve, so A is in 3rd quadrant, i.e. 180 ° <A ° < 270 ° and 0 ° <B ° < 90 ° ( A + B ) ° must be in the 3rd quadrant as tan (A + B ) ° is +ve.

So A + B = 225

tanA ° + tanB °

1 − tanA ° tanB °






Question:

Given that cosy = sin ( x + y ) , show that tany = secx − tanx.

Solution:

cosy = sin ( x + y )⇒ cosy = sinxcosy + cosxsiny

Divide throughout by cosxcosy

⇒ secx = tanx + tany

⇒ tany = secx − tanx






Question:

Given that cotA = and cot (A + B ) = 2 , find the value of cot B.1

4

Solution:

cot ( A + B ) = 2

⇒ tan (A + B ) =

⇒ =

But as cotA = , then tanA = 4.

So =

⇒ 8 + 2tanB = 1 − 4tanB

⇒ 6tanB = − 7

⇒ tanB = −

So cot B = = −

1

2

tanA + tanB

1 − tanA tanB

1

2

1

4

4 + tanB

1 − 4tanB

1

2

7

6

1

tanB

6

7






Question:

Given that tan x + = , show that tanx = 8 − 5√ 3.

π

3

1

2

Solution:

tan x + =

⇒ =

⇒ = tan = √ 3

⇒ 2tanx + 2 √ 3 = 1 − √ 3tanx

⇒ ( 2 + √ 3 ) tanx = 1 − 2√ 3

⇒ tanx = = = = 8 − 5√ 3

π

3

1

2

tanx + tan π

3

1 − tanx tan π

3

1

2

tanx + √ 3

1 − √ 3tanx

1

2

π

3

1 − 2 √ 3

2 + √ 3

( 1 − 2 √ 3 ) ( 2 − √ 3 )

( 2 + √ 3 ) ( 2 − √ 3 )

2 − 4 √ 3 − √ 3 + 6

1




Solutionbank Edexcel AS and A Level Modular Mathematics Exercise B, Question 1

Question:

Write the following expressions as a single trigonometric ratio: (a) 2sin10 ° cos10 °

(b) 1 − 2sin2 25 °

(c) cos2 40 ° − sin2 40 °

(d)

(e)

(f) 6cos2 30 ° − 3

(g)

(h) cos2 − sin2

2 tan5 °

1 − tan2 5 °

1

2sin ( 24 ) ° cos ( 24 ) °1

2

1

2

sin8 °

sec8 °

π

16

π

16

Solution:

(a) 2sin10 ° cos10 ° = sin20 ° (using 2sinAcosA ≡ sin2A)

(b) 1 − 2sin2 25 ° = cos50 ° (using cos2A ≡ 1 − 2sin2 A)

(c) cos2 40 ° − sin2 40 ° = cos80 ° (using cos2A ≡ cos2 A − sin2 A)

(d) = tan10 ° (using tan2A≡ )2 tan5 °

1 − tan2 5 °

2tanA

1 − tan2 A


3/9/2013file://C:\Users\Buba\kaz\ouba\c3_7_b_1.html



(e) = = cosec49 °

(f) 6cos2 30 ° − 3 = 3 ( 2cos2 30 ° − 1 ) = 3cos60 °

(g) = sin8 ° cos8 ° = ( 2sin8 ° cos8 ° ) = sin16 °

(h) cos2 − sin2 = cos = cos

1

2sin ( 24 ) ° cos ( 24 ) °1

2

1

2

1

sin49 °

sin8 °

sec8 °

1

2

1

2

π

16

π

16

2π

16

π

8






Question:

Without using your calculator find the exact values of:

(a) 2sin 22 ° cos 22 °

(b) 2cos2 15 ° − 1

(c) ( sin75 ° − cos75 ° ) 2

(d)

1

2

1

2

2tan π

8

1 − tan2π

8

Solution:

(a) 2sin 22 ° cos 22 ° = sin2 × 22 ° = sin45 ° =

(b) 2cos2 15 ° − 1 = cos ( 2 × 15 ° ) = cos30 ° =

(c) ( sin75 ° − cos75 ° ) 2 = sin2 75 ° + cos2 75 ° − 2sin75 ° cos75 °= 1 − sin ( 2 × 75 ° ) ( sin2 75 ° + cos2 75 ° = 1 )= 1 − sin150 °

= 1 −

=

(d) = tan 2 × = tan = 1

1

2

1

2

1

2

√ 2

2

√ 3

2

1

2

1

2

2 tan π

8

1 − tan2π

8

π

8

π

4





Question:

Write the following in their simplest form, involving only one trigonometric function:

(a) cos2 3θ − sin2 3θ

(b) 6sin2θcos2θ

(c)

(d) 2 − 4sin2

(e) \ 1 + cos2θ

(f) sin2θcos2 θ

(g) 4sinθcosθcos2θ

(h)

(i) sin4θ − 2sin2

θcos2 θ + cos4θ

2tan θ

2

1 − tan2θ

2

θ

2

tanθ

sec2 θ − 2

Solution:

(a) cos2 3θ − sin2 3θ = cos ( 2 × 3θ ) = cos6θ

(b) 6sin2θcos2θ = 3 ( 2sin2θcos2θ ) = 3sin ( 2 × 2θ ) = 3sin4θ

(c) = tan 2 × = tanθ

2tan θ

2

1 − tan2θ

2

θ

2





(d) 2 − 4sin2 θ = 2 1 − 2sin2 θ = 2cos 2 × θ

= 2cosθ

(e) \ 1 + cos2θ = \ 1 + ( 2cos2 θ − 1 ) = \ 2cos2 θ = √ 2cosθ

(f) sin2θcos2 θ = ( 4sin2 θcos2 θ ) = ( 2sinθcosθ ) 2 = sin2 2θ

(g) 4sinθcosθcos2θ = 2 ( 2sinθcosθ ) cos2θ = 2sin2θcos2θ = sin4θ ( sin2A = 2sinAcosA with A = 2θ )

(h) =

=

= −

= −

= − tan2θ

(i) cos4θ − 2sin2

θcos2 θ + sin4θ = ( cos2 θ − sin2

θ ) 2 = cos2 2θ

1

2

1

2

1

2

1

4

1

4

1

4

tanθ

sec2 θ − 2

tanθ

( 1 + tan2 θ ) − 2

tanθ

tan2θ − 1

tanθ

1 − tan2 θ

1

2

2 tanθ

1 − tan2 θ

1

2






Question:

Given that cosx = , find the exact value of cos 2x.1

4

Solution:

cos2x = 2cos2 x − 1 = 2 2 − 1 = − 1 = −

1

4

1

8

7

8






Question:

Find the possible values of sin θ when cos2θ = .23

25

Solution:

cos2θ = 1 − 2sin2 θ

So = 1 − 2sin2 θ

⇒ 2sin2θ = 1 − =

⇒ sin2θ =

⇒ sinθ = ±

23

25

23

25

2

25

1

25

1

5






Question:

Given that cosx + sinx = m and cosx − sinx = n, where m and n are constants, write down, in terms of m and n, the value of cos 2x.

Solution:

cosx + sinx = m cosx − sinx = n Multiply the equations: ( cosx + sinx ) ( cosx − sinx ) = mn ⇒ cos2 x − sin2 x = mn

⇒ cos2x = mn






Question:

Given that tanθ = , and that θ is acute:

(a) Find the exact value of (i) tan 2θ (ii) sin 2θ (iii) cos 2θ

(b) Deduce the value of sin 4θ.

3

4

Solution:

The hypotenuse is 5,

so sinθ = , cosθ = , tanθ =

(a) (i) tan2θ= = = = × =

(ii) sin2θ = 2sinθcosθ = 2 × × =

(iii) cos2θ= cos2 θ − sin2θ = − =

(b) sin4θ= 2sin2θcos2θ= 2 × × =

3

5

4

5

3

4

2tanθ

1 − tan2 θ

3

2

1 −9

16

3

2

7

16

3

2

16

7

24

7

3

5

4

5

24

25

16

25

9

25

7

25

24

25

7

25

336

625





Question:

Given that cosA = − , and that A is obtuse:

(a) Find the exact value of (i) cos 2A (ii) sin A (iii) cosec 2A

(b) Show that tan2A = .

1

3

4 √ 2

7

Solution:

(a) (i) cos2A = 2cos2 A − 1 = 2 − 2 − 1 = − 1 = −

(ii) cos2A = 1 − 2sin2 A

⇒ − = 1 − 2sin2 A

⇒ 2sin2 A = 1 + =

⇒ sin2 A =

⇒ sinA = ± ( √ 8 = 2 √ 2 )

but A is in 2nd quadrant ⇒ sin A is +ve.

So sinA =

(iii) cosec2A = = = = − = −

1

3

2

9

7

9

7

9

7

9

16

9

8

9

2 √ 2

3

2 √ 2

3

1

sin2A

1

2sinAcosA1

2 × × ( − )2 √ 2

3

1

3

9

4 √ 2

9 √ 2

8





(b) tan2A = = = × =sin2A

cos2A

−4 √ 2

9

−7

9

− 4 √ 2

9

− 9

7

4 √ 2

7






Question:

Given that π< θ < , find the value of tan when tanθ = .3π

2

θ

2

3

4

Solution:

Using tanθ =

⇒ =

⇒ 3 − 3tan2 = 8tan

⇒ 3tan2 + 8tan − 3 = 0

⇒ 3tan − 1 tan + 3 = 0

so tan = or tan = − 3

but π< θ <

so < <

i.e. is in the 2nd quadrant

So tan is −ve.

⇒ tan = − 3

2tan θ

2

1 − tan2θ

2

3

4

2tan θ

2

1 − tan2θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

1

3

θ

2

3π

2

π

2

θ

2

3π

4

θ

2

θ

2

θ

2






Question:

In △ABC, AB = 4cm, AC = 5cm, ∠ ABC = 2θ and ∠ ACB = θ. Find the value of θ, giving your answer, in degrees, to 1 decimal place.

Solution:

Using sine rule with =

⇒ =

⇒ =

Cancel sinθ as θ ≠ 0 ° or 180°

So 2cosθ =

⇒ cosθ =

So θ = cos− 1 = 51.3 °

sinB

b

sinC

c

sin2θ

5

sinθ

4

2sinθcosθ

5

sinθ

4

5

4

5

8

5

8






Question:

In △PQR, PQ = 3cm, PR = 6cm, QR = 5cm and ∠ QPR = 2θ.

(a) Use the cosine rule to show that cos2θ= .

(b) Hence find the exact value of sin θ.

5

9

Solution:

(a) Using cosine rule with cosP =

cos2θ= = =

(b) Using cos2θ= 1 − 2sin2 θ

= 1 − 2sin2 θ

⇒ 2sin2θ = 1 − =

⇒ sin2θ =

⇒ sinθ = ±

but sin θ cannot be negative for θ in a triangle

so sinθ =

q2 + r2 − p2

2qr

36 + 9 − 25

2 × 6 × 3

20

36

5

9

5

9

5

9

4

9

2

9

√ 2

3

√ 2

3





Question:

The line l, with equation y = x, bisects the angle between the x-axis and the

line y = mx, m > 0. Given that the scales on each axis are the same, and that l makes an angle θ with the x-axis,

(a) write down the value of tan θ.

(b) Show that m = .

3

4

24

7

Solution:

(a) The gradient of line l is , which is tanθ.

So tanθ =

(b) The gradient of y = mx is m, and as y= x bisects the angle between y = mx

and x-axis

3

4

3

4

3

4





m = tan2θ = = = = × =2 tanθ

1 − tan2 θ

2 ×3

4

1 − ( ) 23

4

3

2

7

16

3

2

16

7

24

7




Solutionbank Edexcel AS and A Level Modular Mathematics Exercise C, Question 1

Question:

Prove the following identities:

(a) ≡ cosA − sinA

(b) − ≡ 2cosec2Asin ( B − A )

(c) ≡ tanθ

(d) ≡ sec2θ

(e) 2 ( sin3θcosθ + cos3 θsinθ ) ≡ sin2θ

(f) − ≡ 2

(g) cosecθ − 2cot 2θcosθ ≡ 2sinθ

(h) ≡ tan2

(i) tan −x ≡

cos2A

cosA + sinA

sinB

sinA

cosB

cosA

1 − cos 2θ

sin2θ

sec2 θ

1 − tan2 θ

sin3θ

sinθ

cos3θ

cosθ

secθ − 1

secθ + 1

θ

2

π

4

1 − sin2x

cos 2x

Solution:

(a) L.H.S. ≡

≡

≡

≡ cosA − sinA ≡ R.H.S.

(b) L.H.S. ≡ −

≡

cos2A

cosA + sinA

cos2 A − sin2 A

cosA + sinA

( cosA + sinA ) ( cosA − sinA )

cosA + sinA

sinB

sinA

cosB

cosA

sinBcosA − cosBsinA

sinAcosA


3/9/2013file://C:\Users\Buba\kaz\ouba\c3_7_c_1.html


≡

≡

≡ 2cosec2Asin ( B − A ) ≡ R.H.S.

(c) L.H.S. ≡

≡

≡

≡

≡ tanθ ≡ R.H.S.

(d) L.H.S. ≡

≡

≡ (as tan2 θ = )

≡

≡ sec2θ≡ R.H.S.

(e) L.H.S. ≡ 2 ( sin3θcosθ + cos3 θsinθ )

≡ 2sinθcosθ ( sin2θ + cos2 θ )

≡ sin2θ (since sin2 θ + cos2 θ ≡ 1 )≡ R.H.S.

(f) L.H.S. ≡ −

≡

≡

sin ( B − A )

( 2sinAcosA )1

2

2sin ( B − A )

sin2A

1 − cos 2θ

sin2θ

1 − ( 1 − 2sin2 θ )

2sinθcosθ

2sin2θ

2sinθcosθ

sinθ

cosθ

sec2 θ

1 − tan2 θ

1

cos2 θ ( 1 − tan2 θ )

1

cos2 θ − sin2θ

sin2θ

cos2 θ

1

cos2θ

sin3θ

sinθ

cos3θ

cosθ

sin3θcosθ − cos3θsinθ

sinθcosθ

sin ( 3θ − θ )

sin2θ1

2




≡

≡ 2 ≡ R.H.S.

(g) L.H.S. ≡ cosecθ − 2cot 2θcosθ

≡ cosecθ − 2 cosθ

≡ cosecθ −

≡ −

≡

≡

≡

≡ 2sinθ ≡ R.H.S.

(h)

(i)

sin2θ

sin2θ1

2

cos2θ

sin2θ

2cos2θcosθ

2sinθcosθ

1

sinθ

cos2θ

sinθ

1 − cos2θ

sinθ

1 − ( 1 − 2sin2 θ )

sinθ

2sin2θ

sinθ










Question:

(a) Show that tanθ + cot θ ≡ 2cosec2θ.

(b) Hence find the value of tan75 ° + cot 75 ° .

Solution:

(a) L.H.S. ≡ tanθ + cot θ

≡ +

≡

≡ ( sin2θ + cos2 θ ≡ 1 )

≡

≡ 2cosec2θ ≡ R.H.S.

(b) Use θ = 75 °

⇒ tan75 ° + cot 75 ° = 2cosec150 ° = 2 × = 2 × = 4

sinθ

cosθ

cosθ

sinθ

sin2θ + cos2 θ

sinθcosθ

2

2sinθcosθ

2

sin2θ

1

sin150 °1

1

2





Question:

Solve the following equations, in the interval shown in brackets. Give answers to 1 decimal place where appropriate.

(a) sin2θ = sinθ { 0 ≤ θ ≤ 2π {

(b) cos2θ = 1 − cosθ { − 180 ° < θ ≤ 180 ° {

(c) 3cos2θ = 2cos2 θ { 0 ≤ θ < 360 ° {

(d) sin4θ = cos2θ { 0 ≤ θ ≤ π {

(e) 2tan2y tany = 3 { 0 ≤ y < 360 ° {

(f) 3cosθ − sin − 1 = 0 0 ≤ θ < 720 °

(g) cos2 θ − sin2θ = sin2θ { 0 ≤ θ ≤ π {

(h) 2sinθ = secθ { 0 ≤ θ ≤ 2π {

(i) 2sin2θ = 3 tanθ { 0 ≤ θ < 360 ° {

(j) 2 tanθ = √ 3 ( 1 − tanθ ) ( 1 + tanθ ) { 0 ≤ θ ≤ 2π {

(k) 5sin2θ + 4sinθ = 0 { − 180 ° < θ ≤ 180 ° {

(l) sin2θ = 2sin2θ { − 180 ° < θ ≤ 180 ° {

(m) 4tanθ = tan2θ { 0 ≤ θ < 360 ° {

θ

2

Solution:

(a) sin2θ = sinθ, 0 ≤ θ ≤ 2π ⇒ 2sinθcosθ = sinθ

⇒ 2sinθcosθ − sinθ = 0

⇒ sinθ ( 2cosθ − 1 ) = 0




⇒ sinθ = 0 or cosθ =

Solution set: 0 , , π , , 2π

(b) cos2θ = 1 − cosθ, − 180 ° <θ ≤ 180 °⇒ 2cos2 θ − 1 = 1 − cosθ

⇒ 2cos2 θ + cosθ − 2 = 0

⇒ cosθ =

As < − 1, cosθ =

As cos θ is +ve, θ is in 1st and 4th quadrants.

Calculator solution is cos− 1 = 38.7 ° .

Solutions are ± 38.7 °

(c) 3cos2θ = 2cos2 θ, 0 ≤ θ < 360 °⇒ 3 ( 2cos2 θ − 1 ) = 2cos2 θ

⇒ 6cos2 θ − 3 = 2cos2 θ

⇒ 4cos2 θ = 3

⇒ cos2 θ =

⇒ cosθ = ±

θ will be in all four quadrants. Solution set: 30°, 150°, 210°, 330°

(d) sin4θ = cos2θ, 0 ≤ θ ≤ π ⇒ 2sin2θcos2θ = cos2θ

⇒ cos2θ ( 2sin2θ − 1 ) = 0

⇒ cos2θ = 0 or sin2θ =

cos2θ = 0 in 0 ≤ 2θ ≤ 2π

⇒ 2θ = , ⇒ θ = ,

sin2θ = in 0 ≤ 2θ ≤ 2π

1

2

π

3

5π

3

− 1 ± \ 17

4

− 1 − \ 17

4

− 1 + \ 17

4

− 1 + \ 17

4

3

4

√ 3

2

1

2

π

2

3π

2

π

4

3π

4

1

2




⇒ 2θ = , ⇒ θ = ,

Solution set: , , ,

(e) 2tan2y tany = 3, 0 ≤ y < 360 °

⇒ tany = 3

⇒ 4tan2 y = 3 − 3tan2 y

⇒ 7tan2 y = 3

⇒ tan2 y =

⇒ tany = ± \

y is in all four quadrants.

y = tan− 1 \ , 180 ° + tan− 1 − \ , 180 ° + tan− 1\ , 360 °

+ tan− 1 − \

y = 33.2°, 146.8°, 213.2°, 326.8°

(f) 3cosθ − sin − 1 = 0, 0 ≤ θ ≤ 720 °

⇒ 3 1 − 2sin2 − sin − 1 = 0

⇒ 6sin2 + sin − 2 = 0

⇒ 3sin + 2 2sin − 1 = 0

π

6

5π

6

π

12

5π

12

π

12

π

4

5π

12

3π

4

4tany

1 − tan2 y

3

7

3

7

3

7

3

7

3

7

3

7

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2




⇒ sin = − or sin =

sin = in 0 ≤ ≤ 360 °

⇒ = 30 ° , 150 ° ⇒ θ = 60 ° , 300 °

sin = − in 0 ≤ ≤ 360 °

⇒ = 180 ° − sin− 1 − , 360 ° + sin− 1 − = 221.8°,

318.2° ⇒ θ = 443.6°, 636.4°

Solution set: 60°, 300°, 443.6°, 636.4°

(g) cos2 θ − sin2θ = sin2θ, 0 ≤ θ ≤ π

⇒ cos2 θ − sin2θ = sin2θ

⇒ cos2θ = sin2θ

⇒ tan2θ = 1 (divide both sides by cos 2θ)tan2θ = 1 in 0 ≤ 2θ ≤ 2π

⇒ 2θ = ,

⇒ θ = ,

(h) 2sinθ = secθ, 0 ≤ θ ≤ 2π

⇒ 2sinθ =

⇒ 2sinθcosθ = 1

⇒ sin2θ = 1

θ

2

2

3

θ

2

1

2

θ

2

1

2

θ

2

θ

2

θ

2

2

3

θ

2

θ

2

2

3

2

3

π

4

5π

4

π

8

5π

8

1

cosθ




sin2θ = 1 in 0 ≤ 2θ ≤ 4π

⇒ 2θ = , (see graph)

⇒ θ = ,

(i) 2sin2θ = 3 tanθ, 0 ≤ θ < 360 °

⇒ 4sinθcosθ =

⇒ 4sinθcos2 θ = 3sinθ

⇒ sinθ ( 4cos2 θ − 3 ) = 0

⇒ sinθ = 0 or cos2 θ =

sinθ = 0 ⇒ θ = 0 ° , 180 °

cos2 θ = ⇒ cosθ = ± ⇒ θ = 30°, 150°, 210°, 330°

Solution set: 0°, 30°, 150°, 180°, 210°, 330°

(j) 2 tanθ = √ 3 ( 1 − tanθ ) ( 1 + tanθ ) , 0 ≤ θ ≤ 2π ⇒ 2tanθ = √ 3 ( 1 − tan2 θ )

⇒ √ 3 tan2 θ + 2tanθ − √ 3 = 0

⇒ ( √ 3tanθ − 1 ) ( tanθ + √ 3 ) = 0

⇒ tanθ = or tanθ = − √ 3

tanθ = , 0 ≤ θ ≤ 2π

⇒ θ = tan− 1 , π + tan− 1 = ,

tanθ = − √ 3, 0 ≤ θ ≤ 2π

π

2

5π

2

π

4

5π

4

3sinθ

cosθ

3

4

3

4

√ 3

2

1

√ 3

1

√ 3

1

√ 3

1

√ 3

π

6

7π

6




⇒ θ = π + tan− 1 ( − √ 3 ) , 2π + tan− 1 ( − √ 3 ) = ,

Solution set: , , ,

(k) 5sin2θ + 4sinθ = 0, − 180 ° <θ ≤ 180 °⇒ 10sinθcosθ + 4sinθ = 0

⇒ 2sinθ ( 5cosθ + 2 ) = 0

⇒ sinθ = 0 or cosθ = −

sinθ = 0 ⇒ θ = 0 ° , 180 ° (from graph)

Calculator value for cos− 1 − is 113.6°

⇒ θ = ± 113.6 °Solution set: −113.6°, 0°, 113.6°, 180°

(l) sin2θ = 2sin2θ, − 180 ° <θ ≤ 180 °

⇒ sin2θ = 4sinθcosθ

⇒ sinθ ( sinθ − 4cosθ ) = 0

2π

3

5π

3

π

6

2π

3

7π

6

5π

3

2

5

2

5





⇒ sinθ = 0 or sinθ = 4cosθ

⇒ sinθ = 0 or tanθ = 4

sinθ = 0 ⇒ θ = 0°, 180°

tanθ = 4 ⇒ θ = tan− 1 4, − 180 ° + tan− 1 4 = 76.0°, −104.0°

Solution set: −104.0°, 0°, 76.0°, 180°

(m) 4tanθ = tan2θ, 0 ≤ θ < 360 °

⇒ 4tanθ =

⇒ 2tanθ ( 1 − tan2 θ ) = tanθ

⇒ tanθ ( 2 − 2 tan2 θ − 1 ) = 0

⇒ tanθ ( 1 − 2 tan2 θ ) = 0

⇒ tanθ = 0 or tanθ = ± \

tanθ = 0 ⇒ θ = 0°, 180°

tanθ = ± \ ⇒ θ = 35.3°, 144.7°, 215.3°, 324.7°

Solution set: 0°, 35.3°, 144.7°, 180°, 215.3°, 324.7°

2 tanθ

1 − tan2 θ

1

2

1

2






Question:

Given that p = 2cosθ and q = cos2θ, express q in terms of p.

Solution:

p = 2cosθ ⇒ cosθ =

cos2θ = q Using cos2θ = 2cos2 θ − 1

q = 2 2 − 1

⇒ q = − 1

p

2

p

2

p2

2





Question:

Eliminate θ from the following pairs of equations:

(a) x = cos2 θ, y = 1 − cos2θ

(b) x = tanθ, y = cot 2θ

(c) x = sinθ, y = sin2θ

(d) x = 3cos2θ + 1, y = 2sinθ

Solution:

(a) cos2 θ = x, cos2θ = 1 − yUsing cos2θ = 2cos2 θ − 1

⇒ 1 − y = 2x − 1

⇒ y = 2 − 2x = 2 ( 1 − x ) (any form)

(b) y = cot 2θ ⇒ tan2θ =

x = tanθ

Using tan2θ =

⇒ =

⇒ 2xy = 1 − x2 (any form)

(c) x = sinθ, y = 2sinθcosθ ⇒ y = 2xcosθ

⇒ cosθ =

Using sin2 θ + cos2 θ ≡ 1

⇒ x2 + = 1

⇒ 4x4 + y2 = 4x2 or y2 = 4x2 ( 1 − x2 ) (any form)

1

y

2 tanθ

1 − tan2 θ

1

y2x

1 − x2

y

2x

y2

4x2





(d) x = 3cos2θ + 1 ⇒ cos2θ =

y = 2sinθ ⇒ sinθ =

Using cos2θ = 1 − 2sin2 θ

⇒ = 1 − = 1 −

⇒ 2 ( x − 1 ) = 6 − 3y2 ( × 6 )

⇒ 3y2 = 6 − 2 ( x − 1 ) = 8 − 2x

⇒ y2 = (any form)

x − 1

3

y

2

x − 1

3

2y2

4

y2

2

2 ( 4 − x )

3






Question:

(a) Prove that ( cos2θ− sin2θ ) 2 ≡ 1 − sin4θ.

(b) Use the result to solve, for 0 ≤ θ < π, the equation cos2θ− sin2θ= .

Give your answers in terms of π.

1

√ 2

Solution:

(a) L.H.S. ≡ ( cos2θ− sin2θ ) 2

≡ cos2 2θ − 2sin2θcos2θ+ sin2 2θ

≡ ( cos2 2θ + sin2 2θ ) − ( 2sin2θcos2θ)≡ 1 − sin4θ ( sin2 A + cos2 A ≡ 1 , sin2A≡ 2sinAcosA )≡ R.H.S.

(b) You can use ( cos2θ− sin2θ ) 2 =

but this also solves cos2θ− sin2θ= −

so you need to check your final answers. As ( cos2θ− sin2θ ) 2 ≡ 1 − sin4θ

⇒ = 1 − sin4θ

⇒ sin4θ=

0 ≤ θ < π, so 0 ≤ 4θ < 4π

⇒ 4θ = , , ,

⇒ θ = , , ,

Checking these values in cos2θ− sin2θ=

eliminates , which apply to cos2θ− sin2θ= −

Solutions are ,

1

2

1

√ 2

1

2

1

2

π

6

5π

6

13π

6

17π

6

π

24

5π

24

13π

24

17π

24

1

√ 2

5π

24

13π

24

1

√ 2

π

24

17π

24





Question:

(a) Show that:

(i) sinθ ≡

(ii) cosθ ≡

(b) By writing the following equations as quadratics in tan , solve, in the

interval 0 ≤ θ ≤ 360 ° : (i) sinθ + 2cosθ = 1 (ii) 3cosθ − 4sinθ = 2 Give answers to 1 decimal place.

2 tan θ

2

1 + tan2θ

2

1 − tan2θ

2

1 + tan2θ

2

θ

2

Solution:

(a) (i) R.H.S. ≡

≡

≡ × cos2

≡ 2sin cos

≡ sinθ ( sin2A = 2sinAcosA )≡ L.H.S.

2tan θ

2

1 + tan2θ

2

2 tan θ

2

sec2θ

2

2sin θ

2

cos θ

2

θ

2

θ

2

θ

2




(ii) R.H.S. ≡

≡

≡ cos2 1 − tan2

≡ cos2 − sin2 tan2 =

≡ cosθ ( cos2A = cos2 A − sin2 A )≡ L.H.S.

(b) Let tan =t

(i) sinθ + 2cosθ = 1

⇒ + = 1

⇒ 2t + 2 − 2t2 = 1 + t2

⇒ 3t2 − 2t − 1 = 0

⇒ ( 3t + 1 ) ( t − 1 ) = 0

⇒ tan = − or tan = 1 0 ≤ ≤ 180 °

tan = 1 ⇒ = 45 ° ⇒ θ = 90 °

tan = − ⇒ = 161.56 ° ⇒ θ = 323.1 °

Solution set: 90°, 323.1° (ii) 3cosθ − 4sinθ = 2

⇒ − = 2

⇒ 3 ( 1 − t2 ) − 8t = 2 ( 1 + t2 )

⇒ 5t2 + 8t − 1 = 0

1 − tan2θ

2

1 + tan2θ

2

1 − tan2θ

2

sec2θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

sin2 θ

2

cos2θ

2

θ

2

2t

1 + t2

2 ( 1 − t2 )

1 + t2

θ

2

1

3

θ

2

θ

2

θ

2

θ

2

θ

2

1

3

θ

2

3 ( 1 − t2 )

1 + t24 × 2t

1 + t2





⇒ t =

For tan = 0 ≤ ≤ 180 °

= 6.65 ° ⇒ θ = 13.3 °

For tan = 0 ≤ ≤ 180 °

= 120.2 ° ⇒ θ = 240.4 °

Solution set: 13.3°, 240.4°

− 8 ± \ 84

10

θ

2

− 8 + \ 84

10

θ

2

θ

2

θ

2

− 8 − \ 84

10

θ

2

θ

2





Question:

(a) Using cos2A ≡ 2cos2 A − 1 ≡ 1 − 2sin2 A, show that:

(i) cos2 ≡

(ii) sin2 ≡

(b) Given that cosθ = 0.6, and that θ is acute, write down the values of:

(i) cos

(ii) sin

(iii) tan

(c) Show that cos4 ≡ ( 3 + 4cosA + cos2A )

x

2

1 + cosx

2

x

2

1 − cosx

2

θ

2

θ

2

θ

2

A

2

1

8

Solution:

(a) (i) Using cos2A ≡ 2cos2 A − 1 with A =

⇒ cosx ≡ 2cos2 − 1

⇒ 2cos2 ≡ 1 + cosx

⇒ cos2 ≡

(ii) Using cos2A ≡ 1 − 2sin2 A

⇒ cosx ≡ 1 − 2sin2

⇒ 2sin2 ≡ 1 − cosx

⇒ sin2 ≡

(b) Given that cosθ = 0.6 and θ acute

x

2

x

2

x

2

x

2

1 + cosx

2

x

2

x

2

x

2

1 − cosx

2





(i) using (a) (i) cos2 = = 0.8 =

⇒ cos = = (as acute)

(ii) using (a) (ii) sin2 = = 0.2 =

⇒ sin = \ =

(iii) tan = = × =

(c) Using (a) (i) and squaring

cos4 = 2 =

but using (a) (i) again

cos2 A = ( 1 + cos2A)

So cos4 = = =

θ

2

1.6

2

4

5

θ

2

2

√ 5

2 √ 5

5

θ

2

θ

2

0.4

2

1

5

θ

2

1

5

√ 5

5

θ

2

sin θ

2

cos θ

2

√ 5

5

5

2 √ 5

1

2

A

2

1 + cosA

2

1 + 2cosA + cos2 A

4

1

2

A

2

1 + 2cosA + ( 1 + cos2A)1

2

42 + 4cosA + 1 + cos2A

8

3 + 4cosA + cos2A

8





Question:

(a) Show that 3cos2 x − sin2 x ≡ 1 + 2cos2x.

(b) Hence sketch, for −π ≤ x ≤ π, the graph of y = 3cos2 x − sin2 x, showing the coordinates of points where the curve meets the axes.

Solution:

(a) R.H.S. ≡ 1 + 2cos2x ≡ 1 + 2 ( cos2 x − sin2 x )

≡ 1 + 2cos2 x − 2sin2 x

≡ cos2 x + sin2 x + 2cos2 x − 2sin2 x (using sin2 x + cos2 x ≡ 1)

≡ 3cos2 x − sin2 x ≡ L.H.S.

(b) y = 3cos2 x − sin2 xis the same as y = 1 + 2cos2x Using your work on transformations this curve is the result of

(i) stretching y = cosx by scale factor in the x direction, then

(ii) stretching the result by scale factor 2 in the y direction, then (iii) translating by 1 in the +ve y direction.

The curve crosses y-axis at (0, 3).

1

2





It crosses x-axis where y = 0 i.e. where 1 + 2cos2x= 0 − π ≤ x ≤ π

⇒ cos2x = − − 2π ≤ 2x ≤ 2π

So 2x = − , , ,

⇒ x = , − , ,

1

2

4π

3

− 2π

3

2π

3

4π

3

− 2π

3

π

3

π

3

2π

3






Question:

(a) Express 2cos2 − 4sin2 in the form acosθ + b, where a and b are

constants.

(b) Hence solve 2cos2 − 4sin2 = − 3, in the interval 0 ≤ θ < 360 ° ,

giving answers to 1 decimal place.

θ

2

θ

2

θ

2

θ

2

Solution:

(a) cos2 = , sin2 =

So 2cos2 − 4sin2 = ( 1 + cosθ ) − 2 ( 1 − cosθ ) = 3cosθ − 1

(b) Hence solve 3cosθ − 1 = − 3, 0 ≤ θ < 360 °⇒ 3cosθ = − 2

⇒ cosθ = −

As cos θ is − ve, θ is in 2nd and 3rd quadrants.

Calculator value is cos− 1 − = 131.8 °

Solutions are 131.8°, 360° − 131.8° = 131.8° , 228.2°

θ

2

1 + cosθ

2

θ

2

1 − cosθ

2

θ

2

θ

2

2

3

2

3





Question:

(a) Use the identity sin2 A + cos2 A ≡ 1 to show that sin4 A + cos4 A ≡

( 2 − sin2 2A ) .

(b) Deduce that sin4 A + cos4 A ≡ ( 3 + cos4A ) .

(c) Hence solve 8sin4θ + 8cos4

θ = 7, for 0 < θ < π.

1

2

1

4

Solution:

(a) As sin2 A + cos2 A ≡ 1so ( sin2 A + cos2 A ) 2 ≡ 1

⇒ sin4 A + cos4 A + 2sin2 Acos2 A ≡ 1

⇒ sin4 A + cos4 A ≡ 1 − 2sin2 Acos2 A

≡ 1 − ( 4sin2 Acos2 A )

≡ 1 − ( 2sinAcosA ) 2

≡ 1 − sin2 2A

≡ ( 2 − sin2 2A )

(b) As cos2A ≡ 1 − 2sin2 Aso cos4A ≡ 1 − 2sin2 2A

so sin2 2A ≡

⇒ from (a) sin4 A + cos4 A ≡ 2 − ≡

≡ 3 + cos4A

1

2

1

2

1

2

1

2

1 − cos4A

2

1

2

1 − cos4A

2

1

2

4 − 1 + cos4A

2

1

4





(c) Using part (b) 8sin4

θ + 8cos4 θ = 7

⇒ 8 × ( 3 + cos4θ ) = 7

⇒ 3 + cos4θ =

⇒ cos4θ =

Solve cos4θ = in 0 < 4θ < 4π

⇒ 4θ = , , ,

⇒ θ = , , ,

1

4

7

2

1

2

1

2

π

3

5π

3

7π

3

11π

3

π

12

5π

12

7π

12

11π

12






Question:

(a) By expanding cos ( 2A + A ) show that cos3A ≡ 4cos3 A − 3cosA.

(b) Hence solve 8cos3 θ − 6cosθ − 1 = 0, for { 0 ≤ θ ≤ 360 ° { .

Solution:

(a) cos ( 2A + A ) ≡ cos2AcosA − sin2AsinA ≡ ( 2cos2 A − 1 ) cosA − ( 2sinAcosA ) sinA

≡ 2cos3 A − cosA − 2sin2 AcosA

≡ 2cos3 A − cosA − 2 ( 1 − cos2 A ) cosA

≡ 2cos3 A − cosA − 2cosA + 2cos3 A

≡ 4cos3 A − 3cosA

(b) 8cos3 θ − 6cosθ − 1 = 0 0 ≤ θ ≤ 360 °⇒ 2 ( 4cos3 θ − 3cosθ ) − 1 = 0

⇒ 2cos3θ − 1 = 0 [using part (a)]

⇒ cos3θ =

Solve cos3θ = in 0 ≤ 3θ ≤ 1080 °

⇒ 3θ = 60 ° , 300°, 420°, 660°, 780°, 1020°

⇒ θ = 20 ° , 100°, 140°, 220°, 260°, 340°

1

2

1

2





Question:

(a) Show that tan3θ ≡ .

(b) Given that θ is acute such that cosθ = , show that tan3θ = .

3 tanθ − tan3θ

1 − 3tan2 θ

1

3

10 √ 2

23

Solution:

(a) tan3θ ≡ tan 2θ + θ ≡

Numerator = + tanθ ≡ ≡

Denominator = 1 − tanθ ≡ ≡

So tan3θ ≡ × ≡

(b) Draw a right-angled triangle.

Using Pythagoras' theorem x2 = 9 − 1 = 8 So x = 2 √ 2 So tanθ = 2 √ 2 Using part (a)

tan2θ+ tanθ

1 − tan2θ tanθ

2 tanθ

1 − tan2 θ

2tanθ + tanθ − tan3θ

1 − tan2 θ

3tanθ − tan3θ

1 − tan2 θ

2 tanθ

1 − tan2 θ

1 − tan2 θ − 2tan2 θ

1 − tan2 θ

1 − 3tan2 θ

1 − tan2 θ

3tanθ − tan3 θ

1 − tan2 θ

1 − tan2 θ

1 − 3 tan2 θ

3 tanθ − tan3θ

1 − 3tan2 θ





tan3θ = = = =3 ( 2 √ 2 ) − ( 2√ 2 ) 3

1 − 3 ( 2√ 2 ) 26 √ 2 − 16√ 2

1 − 24

− 10√ 2

− 23

10 √ 2

23




Solutionbank Edexcel AS and A Level Modular Mathematics Exercise D, Question 1


Question:

Given that 5sinθ + 12cosθ ≡ Rsin ( θ + α ) , find the value of R, R > 0, and the value of tan α.

Solution:

5sinθ + 12cosθ ≡ Rsinθcosα + Rcosθsinα Comparing sinθ : Rcosα = 5 Comparing cosθ Rsinα = 12 Divide the equations:

= ⇒ tanα = 2

Square and add the equations: R2cos2 α + R2sin2

α = 52 + 122

R2 ( cos2 α + sin2α ) = 132

R = 13 since cos2 α + sin2

α ≡ 1

sinα

cosα

12

5

2

5


3/9/2013file://C:\Users\Buba\kaz\ouba\c3_7_d_1.html




Question:

Given that √ 3sinθ + √ 6cosθ ≡ 3cos (θ − α ) , where 0 <α < 90 ° , find the value of α to the nearest 0.1°.

Solution:

√ 3sinθ + √ 6cosθ ≡ 3cosθcosα + 3sinθsinαComparing sinθ : √ 3 = 3sinα �

Comparing cosθ : √ 6 = 3cosα �

Divide � by �:

tanα = =

So α = 35.3 °

√ 3

√ 6

1

√ 2






Question:

Given that 2sinθ − √ 5cosθ ≡ − 3cos (θ + α ) , where 0 <α < 90 ° , find the value of α to the nearest 0.1°.

Solution:

2sinθ − √ 5cosθ ≡ − 3cosθcosα + 3sinθsinα Comparing sinθ : 2 = 3sinα � Comparing cosθ : + √ 5 = + 3cosα � Divide � by �:

tanα =

So α = 41.8 °

2

√ 5






Question:

Show that:

(a) cosθ + sinθ ≡ √ 2sin θ +

(b) √ 3sin2θ − cos2θ ≡ 2sin 2θ −

π

4

π

6

Solution:

(a) R.H.S. ≡ √ 2sin θ +

≡ √ 2 sinθcos + cosθsin

≡ √ 2 sinθ × + cosθ ×

≡ sinθ + cosθ ≡ L.H.S.

(b) R.H.S. ≡ 2sin 2θ −

≡ 2 sin2θcos − cos2θsin

≡ 2 sin2θ × − cos2θ ×

≡ √ 3sin2θ − cos2θ ≡ L.H.S.

π

4

π

4

π

4

1

√ 2

1

√ 2

π

6

π

6

π

6

√ 3

2

1

2






Question:

Prove that cos2θ− √ 3sin2θ ≡ 2cos 2θ + ≡ − 2sin 2θ − .

π

3

π

6

Solution:

Let cos2θ − √ 3sin2θ ≡ Rcos ( 2θ + α ) ≡ Rcos2θcosα − Rsin2θsinα Compare cos2θ : Rcosα = 1 � Compare sin2θ : Rsinα = √ 3 � Divide � by �:

tanα = √ 3 ⇒ α =

Square and add equations: R2 = 1 + 3 = 4

⇒ R = 2

So cos2θ − √ 3sin2θ ≡ 2cos 2θ +

So cos2θ − √ 3sin2θ ≡ 2cos 2θ + ≡ − 2sin 2θ −

π

3

π

3

cos ( 2θ+ )π

3≡ cos2θcos − sin2θsin

π

3

π

3

≡ cos2θ× − sin2θ×1

2

√ 3

2

≡ cos2θsin − sin2θcos π

6

π

6

≡ − ( sin2θcos − cos2θsin )π

6

π

6

≡ − sin ( 2θ− )π

6

π

3

π

6





Question:

Give all angles to the nearest 0.1° and non-exact values of R in surd form. Find the value of R, where R> 0, and the value of α, where 0 <α < 90 ° , in each of the following cases:

(a) sinθ + 3cosθ ≡ Rsin ( θ + α )

(b) 3sinθ − 4cosθ ≡ Rsin ( θ − α )

(c) 2cosθ + 7sinθ ≡ Rcos (θ − α )

(d) cos2θ − 2sin2θ ≡ Rcos ( 2θ + α )

Solution:

(a) sinθ + 3cosθ ≡ Rsinθcosα + Rcosθsinα Compare sinθ : Rcosα = 1 �

Compare cosθ : Rsinα = 3 �

Dividing � by �: tanα = 3 ⇒ α = 71.6 °Square and add equations: R2 = 32 + 12

⇒ R = \ 10

(b) 3sinθ − 4cosθ ≡ Rsinθcosα − Rcosθsinα Compare sinθ : Rcosα = 3 �

Compare cosθ : Rsinα = 4 �

Divide � by �:

tanα = ⇒ α = 53.1 °

Square and add equations: R2 = 32 + 42

⇒ R = 5

(c) 2cosθ + 7sinθ ≡ Rcosθcosα + Rsinθsinα Compare cosθ : Rcosα = 2 �

Compare sinθ : Rsinα = 7 �

Divide � by �:

tanα = ⇒ α = 74.1 °

4

3

7

2





Square and add equations: R2 = 22 + 72 = 53 ⇒ R = \ 53

(d) cos2θ − 2sin2θ ≡ Rcos2θcosα − Rsin2θsinα Compare cos2θ : Rcosα = 1 �

Compare sin2θ : Rsinα = 2 �

Divide � by �: tanα = 2 ⇒ α = 63.4 °Square and add equations: R2 = 12 + 22 = 5 ⇒ R = √ 5





Question:

(a) Show that cosθ − √ 3sinθ can be written in the form Rcos (θ + α ) , with

R > 0 and 0 <α < .

(b) Hence sketch the graph of y = cosθ − √ 3sinθ, 0 < α < 2π, giving the coordinates of points of intersection with the axes.

π

2

Solution:

(a) Let cosθ − √ 3sinθ ≡ Rcos (θ + α ) ≡ Rcosθcosα − Rsinθsinα Compare cosθ : Rcosα = 1 � Compare sinθ : Rsinα = √ 3 �

Divide � by �: tanα = √ 3 ⇒ α =

Square and add: R2 = 1 + 3 = 4 ⇒ R = 2

So cosθ − √ 3sinθ ≡ 2cos θ +

(b) This is the graph of y = cosθ, translated by to the left and then stretched in

the y direction by scale factor 2.

Meets y-axis at (0, 1)

π

3

π

3

π

3





Meets x-axis at , 0 , , 0

π

6

7π

6






Question:

(a) Show that 3sin3θ − 4cos3θ can be written in the form Rsin ( 3θ − α ) , with R > 0 and 0 <α < 90 ° .

(b) Deduce the minimum value of 3sin3θ − 4cos3θ and work out the smallest positive value of θ (to the nearest 0.1°) at which it occurs.

Solution:

(a) Let 3sin3θ − 4cos3θ ≡ Rsin ( 3θ − α ) ≡ Rsin3θcosα − Rcos3θsinα Compare sin3θ : Rcosα = 3 �

Compare cos3θ : Rsinα = 4 �

Divide � by �: tanα = ⇒ α = 53.1 °

Square and add: R2 = 32 + 42⇒ R = 5

So 3sin3θ − 4cos3θ ≡ 5sin ( 3θ − 53.1 ° )

(b) Minimum value occurs when sin ( 3θ − 53.1 ° ) = − 1 So minimum value is − 5 To find smallest +ve value of θ solve sin ( 3θ − 53.1 ° ) = − 1 So 3θ − 53.1 ° = 270 ° ⇒ 3θ = 323.1 °

⇒ θ = 107.7 °

4

3





Question:

(a) Show that cos2θ+ sin2θ can be written in the form Rsin ( 2θ + α ) , with

R > 0 and 0 <α < .

(b) Hence solve, in the interval 0 ≤ θ < 2π, the equation cos2θ + sin2θ = 1, giving your answers as rational multiples of π.

π

2

Solution:

(a) Let cos2θ + sin2θ ≡ Rsin ( 2θ + α ) ≡ Rsin2θcosα + Rcos2θsinα Compare cos2θ : Rsinα = 1 �

Compare sin2θ : Rcosα = 1 �

Divide � by �: tanα = 1 ⇒ α =

Square and add: R2 = 12 + 12 = 2 ⇒ R = √ 2

So cos2θ + sin2θ ≡ √ 2sin 2θ +

(b) Solve √ 2sin 2θ + = 1 , 0 ≤ θ < 2π

so sin 2θ + = , ≤ 2θ + <

As sin 2θ + is +ve, 2θ + is in 1st and 2nd quadrants.

Calculator value is sin− 1 =

π

4

π

4

π

4

π

4

1

√ 2

π

4

π

4

17π

4

π

4

π

4

1

√ 2

π

4





So 2θ + = , , ,

⇒ 2θ = 0, , 2π,

⇒ θ = 0, , π,

π

4

π

4

3π

4

9π

4

11π

4

π

2

5π

2

π

4

5π

4






Question:

(a) Express 7cosθ − 24sinθ in the form Rcos (θ + α ) , with R > 0 and 0 < α < 90 ° . Give α to the nearest 0.1°.

(b) The graph of y= 7cosθ − 24sinθ meets the y-axis at P. State the coordinates of P.

(c) Write down the maximum and minimum values of 7cosθ − 24sinθ.

(d) Deduce the number of solutions, in the interval 0 <θ < 360 ° , of the following equations: (i) 7cosθ − 24sinθ = 15 (ii) 7cosθ − 24sinθ = 26 (iii) 7cosθ − 24sinθ = − 25

Solution:

(a) Let 7cosθ − 24sinθ ≡ Rcos (θ + α ) ≡ Rcosθcosα − Rsinθsinα Compare cosθ : Rcosα = 7 �


Divide � by �: tanα = ⇒ α = 73.7 °

Square and add: R2 = 242 + 72⇒ R = 25

So 7cosθ − 24sinθ ≡ 25cos (θ + 73.7 ° )

(b) Graph meets y-axis where θ= 0, i.e. y = 7cos0 ° − 24sin0 ° = 7 so coordinates are (0, 7)

(c) Maximum value of 25cos (θ + 73.7 ° ) is when cos (θ + 73.7 ° ) = 1 So maximum is 25 Minimum value is 25 ( − 1 ) = − 25

(d) (i) The line y= 15 will meet the graph twice in 0 <θ < 360 ° , so there are 2 solutions. (ii) As the maximum value is 25 it can never be 26, so there are 0 solutions. (iii) As − 25 is a minimum, line y= − 25 only meets curve once, so only 1 solution.

24

7









Question:

(a) Express 5sin2 θ − 3cos2 θ + 6sinθcosθ in the form asin2θ + bcos2θ + c, where a, b and c are constants.

(b) Hence find the maximum and minimum values of 5sin2

θ − 3cos2 θ + 6sinθcosθ.

Solution:

(a) As sin2 θ = and cos2 θ =

so 5sin2 θ − 3cos2 θ + 6sinθcosθ

≡ 5 − 3 + 3 ( 2sinθcosθ )

≡ − cos2θ − − cos2θ + 3sin2θ

≡ 1 − 4cos2θ + 3sin2θ

(b) Write 3sin2θ − 4cos2θ in the form Rsin ( 2θ − α )The maximum value of Rsin ( 2θ − α ) is R The minimum value of Rsin ( 2θ − α ) is − R You know that R2 = 32 + 42 so R = 5 So maximum value of 1 − 4cos2θ + 3sin2θ is 1 + 5 = 6 and minimum value of 1 − 4cos2θ + 3sin2θ is 1 − 5 = − 4

1 − cos2θ

2

1 + cos2θ

2

1 − cos2θ

2

1 + cos2θ

2

5

2

5

2

3

2

3

2





Question:

Solve the following equations, in the interval given in brackets. Give all angles to the nearest 0.1°.

(a) 6sinx + 8cosx = 5 √ 3 [0, 360°]

(b) 2cos3θ − 3sin3θ = − 1 [0, 90°]

(c) 8cosθ + 15sinθ = 10 [0, 360°]

(d) 5sin − 12cos = − 6.5 [ − 360 ° , 360°]x

2

x

2

Solution:

(a) Write 6sinx + 8cosx in the form Rsin ( x + α ) , where R> 0, 0 <α < 90 °so 6sinx + 8cosx ≡ Rsinxcosα + Rcosxsinα Compare sinx : Rcosα = 6 �

Compare cosx : Rsinα = 8 �

Divide � by �: tanα = ⇒ α = 53.13 °

R2 = 62 + 82⇒ R = 10

So 6sinx + 8cosx ≡ 10sin (x + 53.13 ° ) Solve 10sin (x + 53.13 ° ) = 5√ 3 , 0 ≤ x ≤ 360 °

so sin (x + 53.13 ° ) =

⇒ x + 53.13 ° = 60 ° , 120°

⇒ x = 6.9 ° , 66.9°

(b) Let 2cos3θ − 3sin3θ ≡ Rcos ( 3θ + α ) ≡ Rcos3θcosα − Rsin3θsinα Compare cos3θ : Rcosα = 2 �


Divide � by �: tanα = ⇒ α = 56.31 °

R2 = 22 + 32⇒ R = \ 13

Solve \ 13cos ( 3θ + 56.31° ) = − 1 , 0 ≤ θ ≤ 90 °

4

3

√ 3

2

3

2




so cos ( 3θ + 56.31 ° ) = − for

56.31 ° ≤ 3θ + 56.31 ° ≤ 326.31 °

⇒ 3θ + 56.31 ° = 106.1 ° , 253.9°

⇒ 3θ = 49.8 ° , 197.6°

⇒ θ = 16.6 ° , 65.9°

(c) Let 8cosθ + 15sinθ ≡ Rcos (θ − α ) ≡ Rcosθcosα + Rsinθsinα Compare cosθ : Rcosα = 8 �


Divide � by �: tanα = ⇒ α = 61.93 °

R2 = 82 + 152⇒ R = 17

Solve 17cos (θ − 61.93 ° ) = 10 , 0 ≤ θ ≤ 360 °

so cos θ − 61.93 ° = , − 61.93 ° ≤ θ − 61.93 ° ≤ 298.1 °

cos− 1 = 53.97 °

So θ − 61.93° = − 53.97° , + 53.97°

1

\ 13

15

8

10

17

10

17





⇒ θ = 8.0 ° , 115.9°

(d) Let 5sin − 12cos ≡ Rsin − α ≡ Rsin cosα − Rcos sinα

Compare sin : Rcosα = 5 �

Compare cos : Rsinα = 12 �

Divide � by �: tanα = ⇒ α = 67.38 °

R = 13

Solve 13sin − 67.38 ° = − 6.5 , − 360 ° ≤ x ≤ 360 °

so sin − 67.38 ° = − , − 247.4 ° ≤

− 67.4 ° ≤ 112.6 °

From quadrant diagram:

− 67.4 ° = − 150 ° , − 30 °

⇒ = − 82.6 ° , 37.4°

⇒ x = − 165.2 ° , 74.8°

x

2

x

2

x

2

x

2

x

2

x

2

x

2

12

5

x

2

x

2

1

2

x

2

x

2

x

2





Question:

Solve the following equations, in the interval given in brackets. Give all angles to the nearest 0.1°.

(a) sinxcosx = 1 − 2.5cos2x [ 0 , 360 ° ]

(b) cot θ + 2 = cosecθ [ 0 < θ < 360,θ ≠ 180 ]

(c) sinθ = 2cosθ − secθ [ 0 , 180 ° ]

(d) √ 2cos θ − + √ 3 − 1 sinθ = 2 [ 0 , 2π ]

π

4

Solution:

(a) sinxcosx = 1 − 2.5cos2x, 0 ≤ x ≤ 360 °

⇒ sin2x = 1 − 2.5cos2x

⇒ sin2x + 5cos2x = 2Let sin2x + 5cos2x ≡ Rsin ( 2x + α ) ≡ Rsin2xcosα + Rcos2xsinα Compare sin2x : Rcosα = 1 �

Compare cos2x : Rsinα = 5 �

Divide � by �: tanα = 5 ⇒ α = tan− 1 5 = 78.7 °

R2 = 52 + 12⇒ R = \ 26

Solve \ 26 sin ( 2x + 78.7 ° ) = 2, 0 ≤ x ≤ 360 °

⇒ sin 2x + 78.7 ° = , 78.7 ° ≤ 2x + 78.7 ° ≤ 798.7 °

1

2

2

\ 26




⇒ 2x + 78.7 ° = 156.9 ° , 383.1°, 516.9°, 743.1°

⇒ 2x = 78.2 ° , 304.4°, 438.2°, 664.4°

⇒ x = 39.1 ° , 152.2°, 219.1°, 332.2°

(b) cot θ + 2 = cosecθ, 0 < θ < 360 ° , θ ≠ 180 °

⇒ + 2 = (as sinθ ≠ 0)

⇒ cosθ + 2sinθ = 1Let cosθ + 2sinθ ≡ Rcos (θ − α ) ≡ Rcosθcosα + Rsinθsinα Compare cosθ : Rcosα = 1 �


Divide � by �: tanα = 2 ⇒ α = 63.43 °

R2 = 22 + 12⇒ R = √ 5

Solve √ 5cos (θ − 63.43 ° ) = 1 , 0 <θ < 360 °

⇒ cos (θ − 63.43 ° ) = , − 63.43 ° <θ − 63.43 ° < 296.6 °

⇒ θ − 63.43 ° = 63.43 °

⇒ θ = 126.9 °

(c) sinθ = 2cosθ − secθ, 0 ≤ θ ≤ 180 °⇒ sinθcosθ = 2cos2 θ − 1 ( × cosθ )

⇒ sin2θ = cos2θ

⇒ tan2θ = 2, 0 ≤ 2θ ≤ 360 °

⇒ 2θ = tan− 12, 180 ° + tan− 12 = 63.43 ° , 243.43°

⇒ θ = 31.7 ° , 121.7°

(d) √ 2cos θ − + ( √ 3 − 1 ) sinθ

≡ √ 2 cosθcos + sinθsin + ( √ 3 − 1 ) sinθ

≡ cosθ + sinθ + √ 3sinθ − sinθ ≡ cosθ + √ 3sinθ

Let cosθ + √ 3sinθ ≡ Rcos (θ − α ) ≡ Rcosθcosα + Rsinθsinα Compare cosθ : Rcosα = 1 �

Compare sinθ : Rsinα = √ 3 �

Divide � by �: tanα = √ 3 ⇒ α =

cosθ

sinθ

1

sinθ

1

√ 5

1

2

π

4

π

4

π

4

π

3





R2 = ( √ 3 ) 2 + 12⇒ R = 2

Solve 2cos θ − = 2 , 0 ≤ θ ≤ 2π

⇒ cos θ − = 1 , − ≤ θ − ≤

⇒ θ − = 0

⇒ θ =

π

3

π

3

π

3

π

3

5π

3

π

3

π

3





Question:

Solve, if possible, in the interval 0 <θ < 360 ° , θ ≠ 180 ° , the equation

= k in the case when k is equal to:

(a) 4

(b) 2

(c) 1

(d) 0

(e) −1 Give all angles to the nearest 0.1°.

4 − 2 √ 2sinθ

1 + cosθ

Solution:

(a) When k = 4 , 4 − 2√ 2sinθ = 4 + 4cosθ ⇒ − 2 √ 2sinθ = 4cosθ

⇒ tanθ = − = − √ 2

θ = 180 ° + tan− 1 ( − √ 2 ) , 360 ° + tan− 1 ( − √ 2 ) = 125.3 ° , 305.3°

(b) When k = 2, 4 − 2√ 2sinθ = 2 + 2cosθ ⇒ 2cosθ + 2 √ 2sinθ = 2

⇒ cosθ + √ 2sinθ = 1Using the ‘R formula’ L.H.S. ≡ √ 3cos ( θ − 54.74° )

4

2 √ 2





Solve √ 3cos ( θ − 54.74° ) = 1

⇒ cos θ − 54.74 ° =

⇒ θ − 54.74 ° = 54.74 °

⇒ θ = 109.5 °

(c) When k = 1, 4 − 2√ 2sinθ = 1 + cosθ ⇒ cosθ + 2 √ 2sinθ = 3

Using the R formula, cosθ + 2 √ 2sinθ ≡ 3cos (θ − 70.53 ° )Solve 3cos (θ − 70.53 ° ) = 3 ⇒ cos (θ − 70.53 ° ) = 1

⇒ θ − 70.53 ° = 0 °

⇒ θ = 70.5 °

(d) When k = 0, 4 − 2√ 2sinθ = 0 ⇒ sinθ = √ 2

No solutions as − 1 ≤ sinθ ≤ 1

(e) When k = − 1, 4 − 2√ 2sinθ = − 1 − cosθ ⇒ cosθ − 2 √ 2sinθ = − 5

Using the R formula, cosθ − 2 √ 2sinθ ≡ 3cos (θ + 70.53 ° )This lies between −3 and +3, so there can be no solutions.

1

√ 3


3/9/2013PhysicsAndMathsTutor.com

file://C:\Users\Buba\kaz\ouba\c3_7_d_l



Question:

Give all angles to the nearest 0.1° and non-exact values of R in surd form. A class were asked to solve 3cosθ = 2 − sinθ for 0 ≤ θ ≤ 360 ° . One student expressed the equation in the form Rcos (θ − α ) = 2 , with R > 0 and 0 < α < 90 ° , and correctly solved the equation.

(a) Find the values of R and α and hence find her solutions. Another student decided to square both sides of the equation and then form a quadratic equation in sin θ.

(b) Show that the correct quadratic equation is 10sin2θ − 4sinθ − 5 = 0.

(c) Solve this equation, for 0 ≤ θ < 360 ° .

(d) Explain why not all of the answers satisfy 3cosθ = 2 − sinθ.

Solution:

(a) Let 3cosθ + sinθ ≡ Rcos (θ − α ) ≡ Rcosθcosα + Rsinθsinα Compare cosθ : Rcosα = 3 �


Divide � by �: tanα = ⇒ α = 18.43 °

R2 = 32 + 12 = 10 ⇒ R = \ 10 = 3.16

Solve \ 10cos (θ − 18.43 ° ) = 2 , 0 ≤ θ ≤ 360 °

⇒ cos (θ − 18.43 ° ) =

⇒ θ − 18.43 ° = 50.77 ° , 309.23°

⇒ θ = 69.2 ° , 327.7°

(b) Squaring 3cosθ = 2 − sinθ gives 9cos2 θ = 4 + sin2 θ − 4sinθ

⇒ 9 ( 1 − sin2 θ ) = 4 + sin2 θ − 4sinθ

⇒ 10sin2 θ − 4sinθ − 5 = 0

(c) 10sin2θ − 4sinθ − 5 = 0

1

3

2

\ 10


3/9/2013PhysicsAndMathsTutor.com



⇒ sinθ =

For sinθ = , sinθ is +ve, so θ is in 1st and 2nd quadrants.

⇒ θ = 69.2 ° , 180 ° − 69.2 ° = 69.2 ° , 110.8°

For sinθ = , sin θ is −ve, so θ is in 3rd and 4th quadrants.

⇒ θ = 180 ° − ( − 32.3 ° ) , 360 ° + ( − 32.3 ° ) = 212.3 ° , 327.7° So solutions of quadratic in (b) are 69.2°, 110.8°, 212.3°, 327.7°

(d) In squaring the equation, you are also including the solutions to 3cosθ = −( 2 − sinθ ) , which when squared produces the same quadratic.

The extra two solutions satisfying this equation.

4 ± \ 216

20

4 + \ 216

20

4 − \ 216

20




Solutionbank Edexcel AS and A Level Modular Mathematics Exercise E, Question 1

Question:

(a) Show that sin (A + B ) + sin ( A − B ) ≡ 2sinAcosB .

(b) Deduce that sinP + sinQ ≡ 2sin cos .

(c) Use part (a) to express the following as the sum of two sines: (i) 2 sin 7θ cos 2θ (ii) 2 sin 12θ cos 5θ

(d) Use the result in (b) to solve, in the interval 0 ≤ θ ≤ 180 ° , sin3θ + sinθ = 0.

(e) Prove that ≡ .

P + Q

2

P − Q

2

sin7θ + sinθ

sin5θ + sin3θ

cos3θ

cosθ

Solution:

(a) sin ( A + B ) + sin ( A − B ) ≡ sinAcosB + cosAsinB + sinAcosB − cosAsinB ≡ 2sinAcosB

(b) Let P = A + B and Q = A − B, so A = , B =

Substitute in (a): sinP + sinQ ≡ 2sin cos

(c) (i) 2sin7θcos2θ≡ sin ( 7θ+ 2θ ) + sin ( 7θ− 2θ ) [from (a)] ≡ sin9θ+ sin5θ

(ii) 2sin12θsin5θ≡ sin ( 12θ+ 5θ ) + sin ( 12θ− 5θ ) ≡ sin17θ+ sin7θ

(d) sin3θ+ sinθ = 0 ⇒ 2sin cos = 0

so 2sin2θcosθ = 0 ⇒ sin2θ= 0 or cosθ = 0

sin2θ= 0 in 0 ≤ 2θ ≤ 360 °⇒ 2θ = 0 ° , 180°, 360°

⇒ θ = 0 ° , 90°, 180°

cosθ = 0 in 0 ≤ θ ≤ 180 ° ⇒ θ = 90 °Solution set: 0°, 90°, 180°

(e)

P + Q

2

P − Q

2

P + Q

2

P − Q

2

3θ + θ

2

3θ − θ

2


3/9/2013file://C:\Users\Buba\kaz\ouba\c3_7_e_1.html








Question:

(a) Show that sin (A + B ) − sin ( A − B ) ≡ 2cosAsinB .

(b) Express the following as the difference of two sines: (i) 2cos5xsin3x (ii) cos2xsinx

(iii) 6cos xsin x

(c) Using the result in (a) show that sinP − sinQ ≡ 2cos sin .

(d) Deduce that sin56 ° − sin34 ° = √ 2sin11 ° .

3

2

1

2

P + Q

2

P − Q

2

Solution:

(a) sin ( A + B ) − sin ( A − B ) ≡ sinAcosB + cosAsinB −( sinAcosB − cosAsinB )

≡ 2cosAsinB

(b) (i) 2cos5xsin3x≡ sin ( 5x + 3x ) − sin ( 5x− 3x ) ≡ sin8x− sin2x

(ii) cos2xsinx ≡ [ sin ( 2x + x ) − sin ( 2x− x ) ] ≡ ( sin3x− sinx )

(iii) 6cos xsin x ≡ 3 sin x + x − sin x − x ≡ 3

( sin2x − sinx )

(c) In (a) let P = A + B and Q = A − B, so A = , B =

So sinP − sinQ ≡ 2cos sin

(d) sin56 ° − sin34 ° = 2cos sin

= 2cos45 ° sin11 ° = 2 × sin11 ° =√ 2sin11 °

1

2

1

2

3

2

1

2

3

2

1

2

3

2

1

2

P + Q

2

P − Q

2

P + Q

2

P − Q

2

56 ° + 34 °

2

56 ° − 34 °

2

1

√ 2






Question:

(a) Show that cos (A + B ) + cos (A − B ) ≡ 2cosAcosB .

(b) Express as a sum of cosines (i) 2cos cos

(ii) 5cos2xcos3x

(c) Show that cosP + cosQ ≡ 2cos cos .

(d) Prove that ≡ tanθ.

5θ

2

θ

2

P + Q

2

P − Q

2

sin3θ − sinθ

cos3θ + cosθ

Solution:

(a) cos (A + B ) + cos (A − B ) ≡ cosAcosB − sinAsinB + cosAcosB + sinAsinB ≡ 2cosAcosB

(b) Hence, using (a),

(i) 2cos cos ≡ cos + + cos − ≡ cos3θ+ cos2θ

(ii) 5cos2xcos3x≡ ( 2cos3xcos2x)

≡ [ cos ( 3x+ 2x ) + cos ( 3x− 2x ) ] ≡

( cos5x+ cosx )

(c) In (a) let P = A + B, Q = A − B, so A = , B =

So cosP + cosQ ≡ 2cos cos

(d)

5θ

2

θ

2

5θ

2

θ

2

5θ

2

θ

2

5

2

5

2

5

2

P + Q

2

P − Q

2

P + Q

2

P − Q

2






Question:

(a) Show that cos (A + B ) − cos (A − B ) ≡ − 2sinAsinB .

(b) Hence show that cosP − cosQ ≡ − 2sin sin .

(c) Deduce that cos2θ − 1 ≡ − 2sin2θ.

(d) Solve, in the interval 0 ≤ θ ≤ 180 ° , cos3θ + sin2θ − cosθ = 0.

P + Q

2

P − Q

2

Solution:

(a) cos (A + B ) − cos (A − B ) ≡ cosAcosB − sinAsinB − ( cosAcosB + sinAsinB )≡ cosAcosB − sinAsinB − cosAcosB − sinAsinB ≡ − 2sinAsinB

(b) Let P = A + B, Q = A − B, so A = , B =

then cosP − cosQ ≡ − 2sin sin

(c) Let P = 2θ, Q = 0 then cos2θ − cos0≡ − 2sinθsinθ ⇒ cos2θ − 1 ≡ − 2sin2

θ

(d) As cos3θ − cosθ ≡ − 2sin sin

cos3θ − cosθ ≡ − 2sin2θsinθ So cos3θ + sin2θ − cosθ = 0 ⇒ sin2θ − 2sin2θsinθ = 0

⇒ sin2θ ( 1 − 2sinθ ) = 0

⇒ sin2θ = 0 or sinθ =

For sinθ = , θ = 30 ° , 150°

For sin2θ = 0, 2θ = 0 ° , 180°, 360° So θ = 0 ° , 90°, 180° Solution set: 0°, 30°, 90°, 150°, 180°

P + Q

2

P − Q

2

P + Q

2

P − Q

2

3θ + θ

2

3θ − θ

2

1

2

1

2





Question:

Express the following as a sum or difference of sines or cosines:

(a) 2sin8xcos2x

(b) cos5xcosx

(c) 3sinxsin7x

(d) cos100 ° cos40 °

(e) 10cos sin

(f) 2sin30 ° cos10 °

3x

2

x

2

Solution:

(a) 2sin8xcos2x ≡ sin ( 8x + 2x ) + sin ( 8x − 2x )≡ sin10x + sin6x [question 1(a)]

(b) cos5xcosx ≡ ( 2cos5xcosx ) [question 3(a)]

≡ [ cos ( 5x + x ) + cos ( 5x − x ) ] ≡

( cos6x + cos4x )

(c) 3sinxsin7x ≡ − ( − 2sin7xsinx ) [question 4(a)]

≡ − [ cos ( 7x + x ) − cos ( 7x − x ) ] ≡ −

( cos8x − cos6x )

(d) cos100 ° cos40 °≡ ( 2cos100 ° cos40 ° )

≡ [ cos ( 100 + 40 ) ° + cos ( 100 − 40 ) ° ] ≡

( cos140 ° + cos60 ° )

1

2

1

2

1

2

3

2

3

2

3

2

1

2

1

2

1

2





(e) 10cos sin ≡ 5 2cos sin [question 2(a)]

≡ 5 sin + − sin − ≡ 5

sin2x − sinx

(f) 2sin30 ° cos10 °≡ sin ( 30 ° + 10 ° ) + sin ( 30 ° − 10 ° )≡ sin40 ° + sin20 °

3x

2

x

2

3x

2

x

2

3x

2

x

2

3x

2

x

2






Question:

Show, without using a calculator, that 2sin82 ° cos37 ° =

√ 3 + √ 2 .

1

2

1

2

1

2

Solution:

2sin82 ° cos37 ° = sin 82 ° + 37 ° + sin 82 ° − 37

°

= sin120 ° + sin45 °= sin60 ° + sin45 °

= +

= √ 3 + √ 2

1

2

1

2

1

2

1

2

1

2

1

2

√ 3

2

√ 2

2

1

2





Question:

Express, in their simplest form, as a product of sines and/or cosines:

(a) sin12x + sin8x

(b) cos (x + 2y ) − cos ( 2y − x )

(c) ( cos4x + cos2x ) sinx

(d) sin95 ° − sin5 °

(e) cos + cos

(f) sin150 ° + sin20 °

π

15

π

12

Solution:

(a) sin12x + sin8x ≡ 2sin cos

≡ 2sin10xcos2x

(b) cos (x + 2y ) − cos ( 2y − x ) ≡ − 2sin sin

≡ − 2sin2ysinx

(c) cos4x + cos2x ≡ 2cos cos

≡ 2cos3xcosx So ( cos4x + cos2x ) sinx ≡ 2cos3xcosxsinx

≡ cos3x ( 2sinxcosx ) ≡ sin2xcos3x

(d) sin95 ° − sin5 °≡ 2cos sin

≡ 2cos50 ° sin45 ° ≡ √ 2cos50 °

12x + 8x

2

12x − 8x

2

( x + 2y ) + ( 2y − x )

2

( x + 2y ) − ( 2y − x )

2

4x + 2x

2

4x − 2x

2

95 ° + 5 °

2

95 ° − 5 °

2





(e) cos + cos ≡ 2cos cos

≡ 2cos cos − ≡ 2cos cos

(f) sin150 ° + sin20 °≡ 2sin cos

≡ 2sin85 ° cos65 °

π

15

π

12

+π

15

π

12

2

−π

15

π

12

2

9π

120

π

120

9π

120

π

120

150 ° + 20 °

2

150 ° − 20 °

2






Question:

Using the identity cosP + cosQ ≡ 2cos cos , show that

cosθ + cos θ + + cos θ + = 0 .

P + Q

2

P − Q

2

2π

3

4π

3

Solution:

cosθ + cos θ + + cos θ +

≡ cos θ + + cosθ + cos θ +

≡ 2cos cos + cos θ +

≡ 2cos θ + cos + cos θ +

≡ 2cos θ + − + cos θ +

≡ − cos θ + + cos θ +

≡ 0

2π

3

4π

3

4π

3

2π

3

( θ + ) + θ 4π

3

2

( θ + ) − θ 4π

3

2

2π

3

2π

3

2π

3

2π

3

2π

3

1

2

2π

3

2π

3

2π

3






Question:

Prove that = √ 3.sin75 ° + sin15 °

cos15 ° − cos75 °

Solution:

sin75 ° + sin15 ° = 2sin ° cos ° = 2sin45 °

cos30 ° cos15 ° − cos75 ° = − ( cos75 ° − cos15 ° )

= − − 2sin ° sin °

= 2sin45 ° sin30 °

So = = cot 30 ° = = √ 3

75 + 15

2

75 − 15

2

75 + 15

2

75 − 15

2

sin75 ° + sin15 °

cos15 ° − cos75 °

2sin45 ° cos30 °

2sin45 ° sin30 °

1

tan30 °





Question:

Solve the following equations:

(a) cos4x = cos2x, for 0 ≤ x ≤ 180 °

(b) sin3θ − sinθ = 0, for 0 ≤ θ ≤ 2π

(c) sin (x + 20 ° ) + sin (x − 10 ° ) = cos15 ° , for 0 ≤ x ≤ 360 °

(d) sin3θ − sinθ = cos2θ, for 0 ≤ θ ≤ 2π

Solution:

(a) cos4x − cos2x = 0

⇒ − 2sin sin = 0

⇒ sin3xsinx = 0

⇒ sin3x = 0 or sinx = 0, 0 ≤ x ≤ 180 °sinx = 0, 0 ≤ x ≤ 180 °⇒ x = 0 ° , 180°

sin3x = 0, 0 ≤ 3x ≤ 540 °⇒ 3x = 0 ° , 180°, 360°, 540°

⇒ x = 0 ° , 60°, 120°, 180°Solution set: 0°, 60°, 120°, 180°

(b) sin3θ − sinθ = 0

⇒ 2cos sin = 0

⇒ 2cos2θsinθ = 0, 0 ≤ θ ≤ 2π

sinθ = 0, 0 ≤ θ ≤ 2π ⇒ θ = 0, π, 2π cos2θ = 0, 0 ≤ 2θ ≤ 4π

⇒ 2θ = , , ,

⇒ θ = , , ,

4x + 2x

2

4x − 2x

2

3θ + θ

2

3θ − θ

2

π

2

3π

2

5π

2

7π

2

π

4

3π

4

5π

4

7π

4





Solution set: 0, , , π, , , 2π

(c) sin x + 20 ° + sin x − 10 ° ≡ 2sin cos

≡ 2sin ( x + 5 ° ) cos15 ° So sin (x + 20 ° ) + sin (x − 10 ° ) = cos15 ° , 0 ≤ x ≤ 360 °⇒ 2sin (x + 5 ° ) = 1

So sin x + 5 ° = , 5 ° ≤ ( x + 5 ° ) ≤ 365 °

⇒ x + 5 ° = 30 ° , 150°

⇒ x = 25 ° , 145°

(d) sin3θ − sinθ ≡ 2cos sin

≡ 2cos2θsinθ So sin3θ − sinθ = cos2θ ⇒ 2cos2θsinθ = cos2θ

⇒ cos2θ ( 2sinθ − 1 ) = 0

⇒ cos2θ = 0 or sinθ =

sinθ = , 0 ≤ θ ≤ 2π

⇒ θ = ,

cos2θ = 0, 0 ≤ 2θ ≤ 4π

⇒ 2θ = , , ,

⇒ θ = , , ,

Solution set: , , , , ,

π

4

3π

4

5π

4

7π

4

x + 20 ° + x − 10 °

2

x + 20 ° − ( x − 10 ° )

2

1

2

3θ + θ

2

3θ − θ

2

1

2

1

2

π

6

5π

6

π

2

3π

2

5π

2

7π

2

π

4

3π

4

5π

4

7π

4

π

6

π

4

3π

4

5π

6

5π

4

7π

4





Question:

Prove the identities

(a) ≡ 4cos5θ

(b) ≡ − cot θ

(c) sin2 ( x + y ) − sin2 ( x − y ) ≡ sin2xsin2y

(d) cosx + 2cos3x + cos5x ≡ 4cos2 xcos3x

sin7θ− sin3θ

sinθcosθ

cos2θ+ cos4θ

sin2θ− sin4θ

Solution:

(a) L.H.S. ≡

≡

≡

≡ 4cos5θ ≡ R.H.S.

(b) L.H.S. ≡

≡

≡

sin7θ − sin3θ

sinθcosθ

2cos ( 7θ+ 3θ ) sin ( 7θ− 3θ )1

2

1

2

( 2sinθcosθ )1

2

2cos5θsin2θ

sin2θ1

2

cos2θ+ cos4θ

sin2θ− sin4θ

2cos ( 4θ+ 2θ ) cos ( 4θ− 2θ )1

2

1

2

2cos ( 2θ+ 4θ ) sin ( 2θ− 4θ )1

2

1

2

2cos3θcosθ

2cos3θsin ( − θ )





≡ [as sin ( −θ ) ≡ − sinθ ]

≡ − cot θ ≡ R.H.S.

(c) L.H.S. ≡ sin2 ( x + y ) − sin2 ( x − y )≡ [ sin ( x + y ) + sin ( x − y ) ] [ sin ( x + y ) − sin ( x − y ) ]

≡ 2sin cos 2cos

sin

≡ ( 2sinxcosy ) ( 2cosxsiny )≡ ( 2sinxcosx ) ( 2sinycosy )≡ sin2xsin2y ≡ R.H.S.

(d) L.H.S. ≡ cosx + 2cos3x + cos5x ≡ cos5x + cosx + 2cos3x

≡ 2cos cos + 2cos3x

≡ 2cos3xcos2x + 2cos3x ≡ 2cos3x ( cos2x + 1 ) ≡ 2cos3x ( 2cos2 x − 1 + 1 ) ( cos2x ≡ 2cos2 x − 1 )≡ 2cos3x × 2cos2 x

≡ 4cos2 xcos3x ≡ R.H.S.

cosθ

− sinθ

x + y + x − y

2

x + y − x + y

2

x + y + x − y

2

x + y − x + y

2

5x + x

2

5x − x

2





Question:

(a) Prove that cosθ + sin2θ − cos3θ ≡ sin2θ ( 1 + 2sinθ ) .

(b) Hence solve, for 0 ≤ θ ≤ 2π, cosθ + sin2θ = cos3θ.

Solution:

(a) L.H.S. ≡ cosθ + sin2θ − cos3θ ≡ − ( cos3θ − cosθ ) + sin2θ

≡ − − 2sin sin + sin2θ

≡ 2sin2θsinθ + sin2θ ≡ sin2θ ( 2sinθ + 1 ) ≡ R.H.S.

(b) So to solve cosθ + sin2θ = cos3θ or cosθ + sin2θ − cos3θ = 0 solve sin2θ ( 1 + 2sinθ ) = 0 [using (a)] Either sin2θ = 0 ⇒ 2θ = 0, π, 2π, 3π, 4π

⇒ θ = 0, , π, , 2π

or sinθ = −

⇒ θ = π − sin− 1 − , 2π + sin − 1 − = π + , 2π −

3θ + θ

2

3θ − θ

2

π

2

3π

2

1

2

1

2

1

2

π

6





= ,

Solution set: 0, , π, , , , 2π

π

6

7π

6

11π

6

π

2

7π

6

3π

2

11π

6




Solutionbank Edexcel AS and A Level Modular Mathematics Exercise F, Question 1


Question:

The lines l1 and l2, with equations y = 2x and 3y = x − 1 respectively, are drawn

on the same set of axes. Given that the scales are the same on both axes and that the angles that l1 and l2 make with the positive x-axis are A and B respectively,

(a) write down the value of tan A and the value of tan B;

(b) without using your calculator, work out the acute angle between l1 and l2.

Solution:

(a) tanA = 2, tanB = since y = x −

(b) The angle required is (A − B).

Using tan (A − B ) = = = = 1

⇒ A − B = 45 °

1

3

1

3

1

3

tanA − tanB

1 + tanA tanB

2 −1

3

1 + 2 ×1

3

5

3

5

3


3/9/2013file://C:\Users\Buba\kaz\ouba\c3_7_f_1.html




Question:

Given that sinx = where x is acute, and that cos (x − y ) = siny , show that tany =

.

1

√ 5

√ 5 + 1

2

Solution:

As cos (x − y ) = siny cosxcosy + sinxsiny = siny �

Draw a right-angled triangle where sinx =

Using Pythagoras' theorem, a2 = ( √ 5 ) 2 − 1 = 4 ⇒ a = 2

So cosx =

Substitute into �:

cosy + siny = siny

⇒ 2cosy + siny = √ 5siny

⇒ 2cosy = siny ( √ 5 − 1 )

⇒ = tany tany =

⇒ tany = = =

1

√ 5

2

√ 5

2

√ 5

1

√ 5

2

√ 5 − 1

siny

cosy

2 ( √ 5 + 1 )

( √ 5 − 1 ) ( √ 5 + 1 )

2 ( √ 5 + 1 )

4

√ 5 + 1

2





Question:

Using tan2θ = with an appropriate value of θ,

(a) show that tan =√ 2 − 1.

(b) Use the result in (a) to find the exact value of tan .

2 tanθ

1 − tan2 θ

π

8

3π

8

Solution:

(a) Using tan2θ = with θ =

⇒ tan =

Let t = tan

So 1 =

⇒ 1 − t2 = 2t

⇒ t2 + 2t − 1 = 0

⇒ t = = = − 1 ± √ 2

As is acute, tan is +ve, so tan =√ 2 − 1

(b) tan = tan + =

= = = =

2 tanθ

1 − tan2 θ

π

8

π

4

2 tan π

8

1 − tan2π

8

π

8

2t

1 − t2

− 2 ± √ 8

2

− 2 ± 2 √ 2

2

π

8

π

8

π

8

3π

8

π

4

π

8

tan + tan π

4

π

8

1 − tan tan π

4

π

8

1 + ( √ 2 − 1 )

1 − ( √ 2 − 1 )

√ 2

2 − √ 2

√ 2 ( 2 + √ 2 )

( 2 − √ 2 ) ( 2 + √ 2 )

√ 2

2





( 2 + √ 2 ) = √ 2 + 1






Question:

In △ABC, AB = 5 cm and AC = 4 cm, ∠ ABC = ( θ − 30 ) ° and

∠ ACB = θ + 30 ° . Using the sine rule, show that tanθ = 3 √ 3.

Solution:

Using =

⇒ =

⇒ 5sin (θ − 30 ) ° = 4sin (θ + 30 ) °

⇒ 5 ( sinθcos30 ° − cosθsin30 ° ) = 4 ( sinθcos30 °+ cosθsin30 ° )⇒ sinθcos30 ° = 9cosθsin30 °

⇒ = 9 = 9 tan30 °

⇒ tanθ = 9 × = 3√ 3

sinB

b

sinC

c

sin ( θ − 30 ) °

4

sin ( θ + 30 ) °

5

sinθ

cosθ

sin30 °

cos30 °

√ 3

3





Question:

Two of the angles, A and B, in △ABC are such that tanA = , tanB = .

(a) Find the exact value of (i) sin ( A + B )

(ii) tan2B

(b) By writing C as 180 ° − (A + B ) , show that cosC = − .

3

4

5

12

33

65

Solution:

(a) Draw right-angled triangles.

sinA = , cosA = sinB = , cosB =

(i) sin ( A + B ) = sinAcosB + cosAsinB

= × + × =

(ii) tan2B = = = = × =

(b) cosC = cos [ 180 ° − ( A + B ) ] = − cos ( A + B )

3

5

4

5

5

13

12

13

3

5

12

13

4

5

5

13

56

65

2 tanB

1 − tan2 B

2 ×5

12

1 − ( ) 25

12

5

6

119

144

5

6

144

119

120

119





= − ( cosAcosB − sinAsinB ) = − × − ×

= −

4

5

12

13

3

5

5

13





Question:

Show that

(a) secθcosecθ ≡ 2cosec2θ

(b) ≡ sec2 x − 1

(c) cot θ − 2cot 2θ ≡ tanθ

(d) cos4 2θ − sin4 2θ ≡ cos4θ

(e) tan +x − tan −x ≡ 2tan2x

(f) sin ( x + y ) sin ( x − y ) ≡ cos2 y − cos2 x

(g) 1 + 2cos2θ + cos4θ ≡ 4cos2 θcos2θ

1 − cos2x

1 + cos2x

π

4

π

4

Solution:

(a) L.H.S. ≡ secθcosecθ

≡ ×

≡

≡


(b) L.H.S. ≡

≡

≡

1

cosθ

1

sinθ

2

2sinθcosθ

2

sin2θ

1 − cos2x

1 + cos2x

1 − ( 1 − 2sin2 x )

1 + ( 2cos2 x − 1 )

2sin2 x

2cos2 x




≡ tan2 x

≡ sec2 x − 1 ( 1 + tan2 x ≡ sec2 x )≡ R.H.S.

(c) L.H.S. ≡ cot θ − 2cot 2θ

≡ −

≡ −

≡

≡

≡ tanθ ≡ R.H.S.

(d) L.H.S. ≡ cos4 2θ − sin4 2θ≡ ( cos2 2θ + sin2 2θ ) ( cos2 2θ − sin2 2θ )≡ ( 1 ) ( cos4θ ) ( cos2 A + sin2 A ≡ 1 ,

cos2 A − sin2 A ≡ cos2A )≡ cos4θ ≡ R.H.S.

(e) L.H.S. ≡ tan +x − tan −x

≡ −

≡

≡

≡

≡ 2

≡ 2tan2x ≡ R.H.S.

(f) R.H.S. ≡ cos2 y − cos2 x

1

tanθ

2

tan2θ

1

tanθ

2 ( 1 − tan2 θ )

2 tanθ

1 − 1 + tan2 θ

tanθ

tan2θ

tanθ

π

4

π

4

1 + tanx

1 − tanx

1 − tanx

1 + tanx

( 1 + tanx ) 2 − ( 1 − tanx ) 2

( 1 − tanx ) ( 1 + tanx )

1 + 2tanx + tan2 x − ( 1 − 2tanx + tan2 x )

1 − tan2 x

4tanx

1 − tan2 x

2tanx

1 − tan2 x





≡ ( cosy + cosx ) ( cosy − cosx )

≡ 2cos cos − 2sin

sin

≡ 2cos cos 2sin sin

[as sin ( −θ ) = sinθ ]

≡ 2sin cos 2sin cos

≡ sin2 sin2

≡ sin ( x + y ) sin ( x − y )≡ L.H.S.

(g) L.H.S. ≡ 1 + 2cos2θ + cos4θ ≡ 1 + 2cos2θ + ( 2cos2 2θ − 1 )≡ 2cos2θ + 2cos2 2θ ≡ 2cos2θ ( 1 + cos2θ )≡ 2cos2θ [ 1 + ( 2cos2 θ − 1 ) ]≡ 4cos2 θcos2θ ≡ R.H.S.

x + y

2

x − y

2

x + y

2

y − x

2

x + y

2

x − y

2

x + y

2

x − y

2

x + y

2

x + y

2

x − y

2

x − y

2

x + y

2

x − y

2





Question:

The angles x and y are acute angles such that sinx = and cosy =

(a) Show that cos2x= − .

(b) Find the value of cos 2y.

(c) Show without using your calculator, that (i) tan ( x + y ) = 7

(ii) x − y =

2

√ 5

3

\ 10

3

5

π

4

Solution:

(a) cos2x ≡ 1 − 2sin2 x = 1 − 2 2 = 1 − = −

(b) cos2y ≡ 2cos2 y − 1 = 2 2 − 1 = 2 − 1 =

(c)

(i) tan (x + y ) = = = = 7

2

√ 5

8

5

3

5

3

\ 10

9

10

4

5

tanx + tany

1 − tanx tany

2 1

3

1 −2

3

7

3

1

3





(ii) tan ( x − y ) = = = 1

As x and y are acute, x − y = (it cannot be )

tanx − tany

1 + tanx tany

5

3

5

3

π

4

5π

4






Question:

Given that sinxcosy = and cosxsiny = ,

(a) show that sin (x + y ) = 5sin (x − y ) . Given also that tany = k, express in terms of k:

(b) tan x

(c) tan 2x

1

2

1

3

Solution:

(a) sin (x + y ) ≡ sinx cosy + cosxsiny = + =

5sin ( x − y ) ≡ 5 ( sinxcosy − cosxsiny ) = 5 − = 5 × =

(b) = =

⇒ =

so tanx = tany = k

(c) tan2x = = =

1

2

1

3

5

6

1

2

1

3

1

6

5

6

sinxcosy

cosxsiny

1

2

1

3

3

2

tanx

tany

3

2

3

2

3

2

2 tanx

1 − tan2 x

3k

1 − k29

4

12k

4 − 9k2





Question:

Solve the following equations in the interval given in brackets:

(a) √ 3sin2θ + 2sin2θ = 1 { 0 ≤ θ ≤ π {

(b) sin3 θcos2θ = sin2θcos3θ { 0 ≤ θ ≤ 2π {

(c) sin (θ + 40 ° ) + sin (θ + 50 ° ) = 0 { 0 ≤ θ ≤ 360 ° {

(d) sin2 = 2sinθ { 0 ≤ θ ≤ 360 ° {

(e) 2sinθ = 1 + 3cosθ { 0 ≤ θ ≤ 360 ° {

(f) cos5θ = cos3θ { 0 ≤ θ ≤ π {

(g) cos2θ = 5sinθ { − π ≤ θ ≤ π { .

θ

2

Solution:

(a) √ 3sin2θ = 1 − 2sin2 θ, 0 ≤ θ ≤ π⇒ √ 3sin2θ = cos2θ

⇒ tan2θ = , 0 ≤ 2θ ≤ 2π

⇒ 2θ = , π + = ,

⇒ θ = ,

(b) sin3θcos2θ − cos3θsin2θ = 0 ⇒ sin ( 3θ − 2θ ) = 0

⇒ sinθ = 0, 0 ≤ θ ≤ 2π

⇒ θ = 0, π, 2π

(c) sin (θ + 40 ° ) + sin (θ + 50 ° ) = 0 , 0 ≤ θ ≤ 360 °

⇒ 2sin cos

1

√ 3

π

6

π

6

π

6

7π

6

π

12

7π

12

( θ + 40 ° ) + ( θ + 50 ° )

2

( θ + 40 ° ) − ( θ + 50 ° )

2




= 0 ⇒ 2sin (θ + 45 ° ) cos ( − 5 ° ) = 0

⇒ sin ( θ + 45 ° ) = 0 , 45 ° ≤ θ + 45 ° ≤ 405 °

⇒ θ + 45 ° = 180 ° , 360°

⇒ θ = 135 ° , 315°

(d) sin2 = 2 2sin cos sinθ = 2sin cos

⇒ sin sin − 4cos = 0

⇒ sin = 0 or sin = 4cos , i.e. tan = 4

For sin = 0 ⇒ = 0 ° , 180 ° ⇒ θ = 0 ° , 360°

For tan = 4 ⇒ = tan− 1 4 = 75.96 ° ⇒ θ = 151.9 °

Solution set: 0°, 151.9°, 360°

(e) 2sinθ − 3cosθ = 1 Let 2sinθ − 3cosθ ≡ Rsin ( θ − α ) ≡ Rsinθcosα − Rcosθsinα ⇒ Rcosα = 2 and Rsinα = 3

⇒ tanα = ( ⇒ α = 56.3 ° ) , R = \ 13

⇒ \ 13sin (θ − 56.3 ° ) = 1

⇒ sin ( θ − 56.3 ° ) =

⇒ θ − 56.3 ° = sin− 1 , 180 ° − sin− 1 = 16.1 ° , 163.9°

⇒ θ = 72.4 ° , 220.2°

(f) cos5θ − cos3θ = 0

⇒ − 2sin sin = 0

⇒ sin4θsinθ = 0, 0 ≤ θ ≤ π

⇒ sinθ = 0 ⇒ θ = 0, π

or sin4θ = 0 ⇒ 4θ = 0, π, 2π, 3π, 4π

⇒ θ = 0, , , , π

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

θ

2

3

2

1

\ 13

1

\ 13

1

\ 13

5θ + 3θ

2

5θ − 3θ

2

π

4

π

2

3π

4





Solution set: 0, , , , π

(g) cos2θ = 5sinθ ⇒ 1 − 2sin2 θ = 5sinθ

⇒ 2sin2θ + 5sinθ − 1 = 0

⇒ sinθ =

As − 1 ≤ sinθ ≤ 1, sinθ =

In radian mode: θ = 0.187, π − 0.187= 0.187, 2.95

π

4

π

2

3π

4

− 5 ± \ 33

4

− 5 + \ 33

4






Question:

The first three terms of an arithmetic series are √ 3cosθ, sin (θ − 30 ° ) and sin θ, where θ is acute. Find the value of θ.

Solution:

As the three values are consecutive terms of an arithmetic progression, sin ( θ − 30 ° ) − √ 3cosθ = sinθ − sin ( θ − 30 ° ) ⇒ 2sin (θ − 30 ° ) = sinθ + √ 3cosθ

⇒ 2 ( sinθcos30 ° − cosθsin30 ° ) = sinθ + √ 3cosθ

⇒ √ 3sinθ − cosθ = sinθ + √ 3cosθ

⇒ sinθ ( √ 3 − 1 ) = cosθ ( √ 3 + 1 )

⇒ tanθ =

Calculator value is θ = tan− 1 = 75 °

No other values as θ is acute.

√ 3 + 1

√ 3 − 1

√ 3 + 1

√ 3 − 1






Question:

Solve, for 0 ≤ θ ≤ 360 ° , cos ( θ + 40 ° ) cos ( θ − 10 ° ) = 0.5 .

Solution:

2cos (θ + 40 ° ) cos (θ − 10 ° ) = 1

⇒ cos θ + 40 ° + θ − 10 ° + cos θ + 40 °

− θ − 10 ° = 1

⇒ cos ( 2θ + 30 ° ) + cos50 ° = 1

⇒ cos ( 2θ + 30 ° ) = 1 − cos50 ° = 0.3572

⇒ 2θ + 30 ° = 69.07 ° , 290.9°, 429.07°, 650.9°

⇒ 2θ = 39.07 ° , 260.9°, 399.07°, 620.9°

⇒ θ = 19.5 ° , 130.5°, 199.5°, 310.5°






Question:

Without using calculus, find the maximum and minimum value of the following expressions. In each case give the smallest positive value of θ at which each occurs.

(a) sinθcos10 ° − cosθsin10 °

(b) cos30 ° cosθ − sin30 ° sinθ

(c) sinθ + cosθ

Solution:

(a) sinθcos10 ° − cosθsin10 ° = sin (θ − 10 ° ) [ sin (A − B ) ]Maximum value = + 1 when θ − 10 ° = 90 ° ⇒ θ = 100 °

Minimum value = − 1 when θ − 10 ° = 270 ° ⇒ θ = 280 °

(b) cos30 ° cosθ − sin30 ° sinθ = cos (θ + 30 ° )Maximum value = + 1 when θ + 30 ° = 360 ° ⇒ θ = 330 °

Minimum value = − 1 when θ + 30 ° = 180 ° ⇒ θ = 150 °

(c) sinθ + cosθ

= √ 2 sinθ + cosθ

= √ 2 ( sinθcos45 ° + cosθsin45 ° )= √ 2sin (θ + 45 ° )

Maximum value =√ 2 when θ+ 45 ° = 90 ° ⇒ θ = 45 °

Minimum value = − √ 2 when θ + 45 ° = 270 ° ⇒ θ = 225 °

1

√ 2

1

√ 2






Question:

(a) Express sinx − √ 3cosx in the form Rsin ( x − α ) , with R > 0 and 0 <α < 90 ° .

(b) Hence sketch the graph of y = sinx − √ 3cosx { − 360 ° ≤ x ≤ 360 ° { , giving the coordinates of all points of intersection with the axes.

Solution:

(a) Let sinx − √ 3cosx ≡ Rsin ( x − α ) ≡ Rsinxcosα − Rcosxsinα R > 0, 0 <α < 90 ° Compare sinx : Rcosα = 1 � Compare cosx : Rsinα = √ 3 �

Divide � by �: tanα = √ 3 ⇒ α = 60 °

R2 = ( √ 3 ) 2 + 12 = 4 ⇒ R = 2

So sinx − √ 3cosx ≡ 2sin (x − 60 ° )

(b) Sketch y = 2sin (x − 60 ° ) by first translating y = sinx by 60° to the right and then stretching the result in the y direction by scale factor 2.

Graph meets y-axis when x = 0, i.e. y = 2sin ( − 60 ° ) = − √ 3 Graph meets x-axis when y = 0, i.e. (− 300 ° , 0), ( − 120° , 0), (60°, 0), 240°, 0)






Question:

Given that 7cos2θ+ 24sin2θ ≡ Rcos ( 2θ − α ) , where R> 0 and 0 <α < ,

find:

(a) the value of R and the value of α, to 2 decimal places

(b) the maximum value of 14cos2 θ + 48sinθcosθ

π

2

Solution:

(a) Let 7cos2θ + 24sin2θ ≡ Rcos ( 2θ − α ) ≡ Rcos2θcosα + Rsin2θsinα

R > 0, 0 <α <

Compare cos2θ : Rcosα = 7 �


Divide � by �: tanα = ⇒ α = 1.29 (1.287)

R2 = 242 + 72⇒ R = 25

So 7cos2θ + 24sin2θ ≡ 25cos ( 2θ − 1.29 )

(b) 14cos2 θ + 48sinθcosθ

≡ 14 + 24 ( 2sinθcosθ )

≡ 7 ( 1 + cos2θ ) + 24sin2θ ≡ 7 + 7cos2θ + 24sin2θ

The maximum value of 7cos2θ + 24sin2θ is 25 [using (a) with cos( 2θ − 1.29 ) = 1 ].

So maximum value of 7 + 7cos2θ + 24sin2θ = 7 + 25 = 32.

π

2

24

7

1 + cos2θ

2






Question:

(a) Given that α is acute and tanα = , prove that

3sin (θ + α ) + 4cos (θ + α ) ≡ 5cosθ

(b) Given that sinx = 0.6 and cosx = − 0.8, evaluate cos (x + 270 ) ° and cos( x + 540 ) ° .

3

4

Solution:

(a) Draw a right-angled triangle and find sin α and cos α.

⇒ sinα = , cosα =

So 3sin (θ + α ) + 4cos (θ + α )≡ 3 ( sinθcosα + cosθsinα ) + 4 ( cosθcosα − sinθsinα )

≡ 3 sinθ + cosθ + 4 cosθ − sinθ

≡ sinθ + cosθ + cosθ − sinθ

≡ cosθ

≡ 5cosθ

(b) cos (x + 270 ) ° ≡ cosx ° cos270 ° − sinx ° sin270 °= ( − 0.8 ) ( 0 ) − ( 0.6 ) ( − 1 ) = 0 + 0.6 = 0.6

cos (x + 540 ) ° ≡ cosx ° cos540 ° − sinx ° sin540 °= ( − 0.8 ) ( − 1 ) − ( 0.6 ) ( 0 ) = 0.8 − 0 = 0.8

3

5

4

5

4

5

3

5

4

5

3

5

12

5

9

5

16

5

12

5

25

5





Question:

(a) Without using a calculator, find the values of: (i) sin40 ° cos10 ° − cos40 ° sin10 °

(ii) cos15 ° − sin15 °

(iii)

(b) Find, to 1 decimal place, the values of x, 0 ≤ x ≤ 360 ° , which satisfy the equation 2sinx = cos (x − 60 )

1

√ 2

1

√ 2

1 − tan15 °

1 + tan15 °

Solution:

(a) (i) sin40 ° cos10 ° − cos40 ° sin10 ° = sin ( 40 ° − 10 ° ) = sin30 ° =

(ii) cos15 ° − sin15 °

= cos45 ° cos15 ° − sin45 ° sin15 ° = cos ( 45 ° + 15 ° ) = cos60 °

=

(iii) =

= tan ( 45 ° − 15 ° ) = tan30 ° =

(b) 2sinx = cos (x − 60 ° )⇒ 2sinx = cosxcos60 ° + sinxsin60 °

⇒ 2sinx = cosx + sinx

⇒ ( 4 − √ 3 ) sinx = cosx

⇒ =

⇒ tanx =

1

2

1

√ 2

1

√ 2

1

2

1 − tan15 °

1 + tan15 °

tan45 ° − tan15 °

1 + tan45 ° tan15 °

√ 3

3

1

2

√ 3

2

sinx

cosx

1

4 − √ 3

1

4 − √ 3





So x = tan− 1 , 180 ° + tan− 1

⇒ x = 23.8 ° , 203.8 °

1

4 − √ 3

1

4 − √ 3






Question:

(a) Prove, by counter example, that the statement ‘sec (A + B ) ≡ secA + secB , for all A and B’ is false.

(b) Prove that tanθ + cot θ ≡ 2cosec2θ, θ ≠ , n∈ ℤ.nπ

2

Solution:

(a) One example is sufficient to disprove a statement. E.g. A= 60 ° , B= 0 °

sec (A + B ) = sec ( 60 ° + 0 ° ) = sec60 ° = = 2

secA = sec60 ° = = 2

secB = sec0 ° = = 1

So sec ( 60 ° + 0 ° )≠ sec60 ° + sec0 °⇒ sin ( A + B ) ≡ secA + secB not true for all values of A, B.

(b) L.H.S. ≡ tanθ + cot θ

≡ +

≡

≡ ( sin2θ + cos2 θ ≡ 1 , sin2θ≡ 2sinθcosθ )

≡


1

cos60 °

1

cos60 °

1

cos0 °

sinθ

cosθ

cosθ

sinθ

sin2θ + cos2 θ

sinθcosθ

1

sin2θ1

2

2

sin2θ








Question:

Using the formula cos (A + B ) ≡ cosAcosB − sinAsinB :

(a) Show that cos (A − B ) − cos (A + B ) ≡ 2sinAsinB .

(b) Hence show that cos2x − cos4x ≡ 2sin3xsinx.

(c) Find all solutions in the range 0 ≤ x ≤ π of the equation cos2x − cos4x = sinx giving all your solutions in multiples of π radians.

Solution:

(a) cos (A + B ) ≡ cosAcosB − sinAsinB

⇒ cos A − B ≡ cosAcos −B − sinAsin − B

≡ cosAcosB + sinAsinB so cos (A + B ) − cos (A − B ) ≡ ( cosAcosB − sinAsinB ) −( cosAcosB + sinAsinB )≡ − 2sinAsinB

(b) Let A + B = 2x, A − B = 4x Add: 2A = 6x ⇒ A = 3x

Subtract: 2B = − 2x ⇒ B = − x Using (a) cos2x − cos4x ≡ − 2sin3xsin ( − x ) ≡ 2sin3xsinx as sin ( −x ) = − sinx

(c) Solve 2sin3xsinx = sinx ⇒ sinx ( 2sin3x − 1 ) = 0

⇒ sinx = 0 or sin3x =

sinx = 0 ⇒ x = 0, π

sin3x = , 0 ≤ 3x ≤ 3π ⇒ 3x = , , ,

⇒ x = , , ,

1

2

1

2

π

6

5π

6

13π

6

17π

6

π

18

5π

18

13π

18

17π

18





Solution set: 0, , , , , ππ

18

5π

18

13π

18

17π

18





Question:

(a) Given that cos (x + 30 ° ) = 3cos (x − 30 ° ) , prove that tanx = − .

(b) (i) Prove that = tanθ.

(ii) Verify that θ = 180 ° is a solution of the equation sin2θ = 2 − 2cos2θ. (iii) Using the result in part (i), or otherwise, find the two other solutions, 0 < θ < 360 ° , of the equation sin2θ = 2 − 2cos2θ.

√ 3

2

1 − cos2θ

sin2θ

Solution:

(a) cos (x + 30 ° ) = 3cos (x − 30 ° )⇒ cosxcos30 ° − sinxsin30 ° = 3 ( cosxcos30 ° + sinxsin30 ° )

⇒ − 2cosxcos30 ° = 4sinxsin30 °

⇒ − 2cosx × = 4sinx ×

⇒ − =

⇒ tanx = −

(b) (i) L.H.S. ≡

≡

≡

≡

≡ tanθ (ii) L.H.S. = sin360 ° = 0 R.H.S. = 2 − 2cos360 ° = 2 − 2 ( 1 ) = 0 �

(iii) Using (i) this is equivalent to solving tanθ = .

√ 3

2

1

2

√ 3

2

sinx

cosx

√ 3

2

1 − cos2θ

sin2θ

1 − ( 1 − 2sin2 θ )

2sinθcosθ

2sin2θ

2sinθcosθ

sinθ

cosθ

1

2





From (i) 1 − cos2θ = sin2θ tanθ So sin2θ = 2 − 2cos2θ ⇒ sin2θ = 2sin2θ tanθ

sin2θ = 0 gives θ = 180 ° , so tanθ = ⇒ θ = 26.6°, 206.6°1

2






Question:

(a) Express 1.5sin2x+ 2cos2x in the form Rsin ( 2x+ α ) , where R > 0 and 0 <α <

, giving your values of R and α to 3 decimal places where appropriate.

(b) Express 3sinxcosx + 4cos2 x in the form asin2x + bcos2x+ c, where a, b and c are constants to be found.

(c) Hence, using your answer to part (a), deduce the maximum value of 3sinxcosx + 4cos2 x.

π

2

Solution:

(a) Let 1.5sin2x+ 2cos2x≡ Rsin ( 2x + α ) ≡ Rsin2xcosα + Rcos2xsinα

R > 0, 0 <α <

Compare sin2x: Rcosα = 1.5 �

Compare cos2x: Rsinα = 2 �

Divide � by �: tanα = ⇒ α = 0.927

R2 = 22 + 1.52⇒ R = 2.5

(b) 3sinxcosx + 4cos2 x ≡ ( 2sinxcosx ) + 4

≡ sin2x+ 2 + 2cos2x≡ sin2x+ 2cos2x+ 2

(c) From part (a) sin2x+ 2cos2x≡ 2.5 sin ( 2x+ 0.927 )

So maximum value of sin2x+ 2cos2x= 2.5 × 1 = 2.5

So maximum value of 3sinxcosx + 4cos2 x = 2.5 + 2 = 4.5

π

2

4

3

3

2

1 + cos2x

2

3

2

3

2

3

2

3

2




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